IN  MEMORIAM 
FLOR1AN  CAJORI 


// 


SECONDARY  ALGEBRA 


BY 


GEORGE  EGBERT  FISHER,  M.A.,  PH.D. 


1 1 

AND 


ISAAC  J.   SCHWATT,  Pn.D. 

ASSISTANT   PROFESSORS  OF  MATHEMATICS  IN  THE 
UNIVERSITY   OF  PENNSYLVANIA 


PHILADELPHIA 

FISHER    AND    SCHWATT 
1900 


COPYRIGHT,  1900, 
BY  FISHER  AND  SCHWATT. 


Nortoooto 

J.  S.  Gushing  &  Co.  -  Berwick  &  Smith 
Norwood  Mass.  U.S.A. 


PREFACE. 

IN  the  preparation  of  this  book,  the  aim  of  the  authors  has 
been  to  give  the  student  a  working  knowledge  of  the  ele- 
mentary processes  of  algebra,  with  a  conviction  of  the  truth 
of  principles  through  illustrations  and  particular  examples. 
Each  principle,  or  method,  is  therefore  first  clearly  illustrated 
by  numerous  and  simple  exercises  worked  in  the  text.  But 
the  student  is  not  left  to  assume  that  the  principles  are  thereby 
proved.  Even  a  beginner  should  not  be  encouraged,  by  text- 
book or  teacher,  to  accept  an  illustrative  example  as  a  proof, 
or  he  will  lose  much  of  the  educational  value  of  the  study. 

Particular  attention  has  been  paid  to  the  grading  of  the 
exercises. 

The  introductory  chapter  extends  the  familiar  processes  of 
arithmetic  to  the  corresponding  processes  of  algebra.  The 
pupil  is  led  by  simple  exercises,  similar  to  those  in  arithmetic, 
to  understand  the  use  of  letters  to  represent  general  and 
unknown  numbers.  Negative  numbers  are  naturally  intro- 
duced in  connection  with  the  extension  of  subtraction  of 
arithmetical  numbers.  The  meaning  and  use  of  positive  and 
negative  numbers,  in  the  fundamental  operations,  are  properly 
emphasized. 

Equations  and  problems  are  distributed  throughout  the 
book.  The  importance  of  equivalent  equations  is  not  over- 
looked, but  is  very  briefly  and  simply  considered  in  Chapter 
IV.  Until  that  chapter  is  reached,  the  solutions  of  equations 

should  be  checked. 

iii 


iv  PREFACE. 

All  the  matter  in  the  book  is  printed  in  large  type,  and 
much  pains  has  been  taken  to  make  the  pages  open  and 
attractive. 

Any  suggestions  from  teachers  and  others  will  be  greatly 
appreciated. 

The  authors  have  much  pleasure  in  expressing  their  satisfac- 
tion with  the  excellence  of  the  mechanical  execution  of  the 
work,  due  to  the  ability  and  painstaking  care  of  Messrs.  J.  S. 
Gushing  &  Co.  and  Messrs.  Berwick  &  Smith,  of  the  Norwood 
Press. 

G.  E.  F. 
I.  J.  S. 

UNIVERSITY  OF  PENNSYLVANIA, 
PHILADELPHIA. 


CONTENTS. 


CHAPTER   I. 

PAGE 

INTRODUCTION     ...........  1 

Genera]  Number 1 

Equations  without  Transposition 9 

Problems 10 

Positive  and  Negative  Numbers .18 

CHAPTER  II. 

FUNDAMENTAL  OPERATIONS  WITH  ALGEBRAIC  NUMBERS  ...  24 

Addition 24 

Subtraction 25 

Multiplication      ..........  27 

Division       .         .         ...         .         .         .         .         .         .30 

Parentheses          ..........  34 

Positive  Integral  Powers     ........  39 

CHAPTER   III. 

FUNDAMENTAL  OPERATIONS   WITH   INTEGRAL  ALGEBRAIC   EXPRES- 
SIONS           42 

Addition  and  Subtraction    ........  44 

Parentheses         ..........  51 

Equations  with  Transposition,  and  Problems      ....  53 

Multiplication                       t 57 

Equations  and  Problems 66 

Division 69 

CHAPTER   IV. 

INTEGRAL  ALGEBRAIC  EQUATIONS      .         .         .         .         .         .         .79 

Equivalent  Equations 81 

Problems 84 

v 


vi  CONTENTS. 

i 

CHAPTER  V. 

PAGE 

TYPE-FORMS  IN  MULTIPLICATION  AND  DIVISION       ....  89 

CHAPTER  VI. 

FACTORS  AND  MULTIPLES  OF  INTEGRAL  ALGEBRAIC  EXPRESSIONS  .  99 

Integral  Algebraic  Factors           .......  99 

Highest  Common  Factors 116 

Lowest  Common  Multiples 118 

Solutions  of  Equations  by  Factoring 127 

CHAPTER   VII. 

FRACTIONS 129 

Reduction  to  Lowest  Terms         .        .         .        .        .        .         .131 

Reduction  to  Lowest  Common  Denominator      ....  133 

Equations  and  Problems 136 

Addition  and  Subtraction   ........  141 

Multiplication      ..........  149 

Division 153 

Complex  Fractions      .........  155 

CHAPTER  VIII. 

FRACTIONAL  EQUATIONS     .                  159 

Problems     .         .         . 162 

CHAPTER  IX. 

LITERAL  EQUATIONS 167 

General  Problems 169 

Interpretation  of  Solutions 172 

CHAPTER   X. 

SIMULTANEOUS  LINEAR  EQUATIONS    .......  177 

Elimination  by  Addition  and  Subtraction  .....  178 

Elimination  by  Comparison 180 

Elimination  by  Substitution        .         .         .         .         .         .         .  182 

Linear  Equations  in  Three  Unknown  Numbers  .        .         .        .183 

Problems  192 


CONTENTS.  vii 
CHAPTER  XI. 

PAGE 

INEQUALITIES .        .  204 

CHAPTER  XII. 

INDETERMINATE  EQUATIONS       .        .        .        .        ,        .        .        .  210 

CHAPTER   XIII. 

INVOLUTION 213 

Powers  of  Binomials 215 

Powers  of  Multinomials      .        .        .        .        .        .        .        ..217 

CHAPTER   XIV. 

EVOLUTION 218 

Hoots  of  Monomials 221 

Square  Roots  of  Multinomials     .         .        .        .        .        .         .  222 

Cube  Roots  of  Multinomials        . 226 

Roots  of  Arithmetical  Numbers 228 

CHAPTER  XV. 

SURDS 235 

Reduction  of  Surds 236 

Addition  and  Subtraction  of  Surds 239 

Reduction  of  Surds  of  Different  Orders  to  Equivalent  Surds  of 

the  Same  Order 240 

Multiplication  of  Surds        . 241 

Division  of  Surds         . 246 

Rationalization 247 

Surd  Factors 248 

Evolution  of  Surd  Expressions 250 

Properties  of  Quadratic  Surds     .......  251 

Square  Roots  of  Simple  Binomial  Surds      .....  252 

Approximate  Values  of  Surd  Numbers 253 

Irrational  Equations    ......                 .  255 


viii  CONTENTS. 

CHAPTER  XVI. 

PAGB 

IMAGINARY  AND  COMPLEX  NUMBERS 257 

Imaginary  Numbers 257 

Complex  Numbers 2(il 

Complex  Factors         .                 .                 263 

CHAPTER   XVII. 

DOCTRINE  OF  EXPONENTS           ........  265 

Zeroth  Powers 266 

Negative  Integral  Powers    ........  267 

Fractional  (Positive  or  Negative)  Powers  .....  269 

CHAPTER  XVIII. 

QUADRATIC  EQUATIONS       .........  277 

Pure  Quadratic  Equations  .         ..         .         .        .        .         .277 

Solution  by  Factoring 280 

Solution  by  Completing  the  Square 282 

Relation  between  the  Roots  and  Coefficients       ....  287 

Nature  of  the  Roots 288 

Irrational  Equations 290 

Higher  Equations 293 

Problems 295 

CHAPTER   XIX. 

SIMULTANEOUS  QUADRATIC  EQUATIONS 300 

Problems 307 

CHAPTER  XX. 

RATIO,  PROPORTION,  AND  VARIATION 311 

Ratio 311 

Proportion 313 

Variation 320 

CHAPTER   XXI. 

PROGRESSIONS     ...........  324 

Arithmetical  Progression     ........  324 

Geometrical  Progression      ........  331 

Harmonical  Progression       ........  339 


CONTENTS.  ix 
CHAPTER   XXII. 

PAGE 

BINOMIAL  THEOREM  FOR  POSITIVE  INTEGRAL  EXPONENT         .         .  342 

CHAPTER   XXIII. 

VARIABLES  AND  LIMITS ,  348 

Variables 348 

Limits          .         .         .         .         .         .         .     -   .         .         .         •  348 

Indeterminate  Fractions 352 

Interpretation  of  Solutions          .......  353 

CHAPTER   XXIV. 

UNDETERMINED  COEFFICIENTS  .         .         .         .         .         .         .         .  358 

Convergent  and  Divergent  Series 358 

Expansion  of  Rational  Fractions 361 

Expansion  of  Surds     .........  364 

Partial  Fractions          .........  364 

CHAPTER  XXV. 

THE  BINOMIAL  THEOREM  FOR  ANY  RATIONAL  EXPONENT        .         .  370 

CHAPTER   XXVI. 

LOGARITHMS        ...........  373 

Principles  of  Logarithms     ........  374 

Systems  of  Logarithms        ........  377 

Properties  of  Common  Logarithms      ......  378 

To  find  the  Logarithm  of  a  Given  Number          .         .        .         .381 

To  find  a  Number  from  its  Logarithm         .....  385 

C  ©logarithms 388 

Applications 389 

Exponential  Equations 393 

Compound  Interest  and  Annuities 395 

TABLE  OF  LOGARITHMS  .         .         .         .         .         .         .  -       i-xviii 


CHAPTER  I. 

INTRODUCTION. 

GENERAL  NUMBER. 

1.  Algebra,  like  Arithmetic,  treats  of  number. 

2.  The  examples 

2     3      23 


77         7         7  1111        11        11 

are  particular  cases  of  the  following  principle  : 

The  sum  of  two  fractions  which  have  a  common  denominator 
is  a  fraction  whose  denominator  is  the  common  denominator,  and 
whose  numerator  is  the  sum  of  the  numerators;  or,  more  briefly 
stated, 

1st  num.       2d  num.  _  1st  num.  +  %d  num. 
com.  den.      com.  den.  com.  den. 

This  principle  can  be  stated  still  more  concisely  by  letting 
letters  stand  for  the  two  numerators  and  the  common  de- 
nominator. 

Let  a  stand  for  1st  num.,  b  for  2d  num.,  and  c  for  com.  den. 
We  then  have 

a      b  _  a  4-  b 
c      c          c 

This  relation  states  by  means  of  letters,  or  symbols,  all  that 
is  contained  in  the  verbal  statement.  The  letters  a,  6,  and  c 
stand  for  the  terms  of  any  two  fractions,  and  therefore  denote 
any  numbers  whatever. 

In  the  first  example  above,  a  =  2,  b  =  3,  c  =  7;  in  the  second, 
a  =  5,  b  =  4,  c  =  11. 

1 


2  ALGEBRA.  [Cii.  I 

3.  In  ordinary  Arithmetic  all  numbers  are  represented  by 
the  Arabic  numerals,  1,  2,  3,  etc.    Each  of  these  symbols  stands 
for  a  definite  number.     The  symbol  7,  for  instance,  stands  for 
a  group  of  seven  units,  the  symbol  5  for  a  group  of  Jive  units. 

But  in  Algebra,  such  symbols  as  a,  6,  x,  y,  are  used  to  repre- 
sent numbers  which  may  have  any  values  whatever,  as  in  Art.  2. 

For  the  sake  of  brevity  we  shall  say  the  number  a,  or  simply 
a,  meaning  thereby  the  number  denoted  by  the  symbol  a. 

4.  The  numbers  represented  by  letters  are,  for  the  sake  of 
distinction,  called  Literal  or  General  Numbers. 

EXERCISES   I. 

If  p  is  the  product  obtained  by  multiplying  a  by  6,  express 
in  symbols  the  following  principles  of  multiplication  : 

1.  The  multiplicand  is  equal  to  the  product  divided  by  the 
multiplier.     Let  p  =  35,  a  —  7,  b  =  5  -,  p  —  24,  a  =  3,  b  —  8. 

2.  The  multiplier  is  equal.  to  the  product  divided   by  the 
multiplicand.     Let  p  =  63,  a  =  9,  b  =  7  ,  p  =  40,  a  —  5,  b  =  8. 

If  q  is  the  quotient  obtained  by  dividing  m  by  n,  express  in 
symbols  the  following  principles  of  division  : 

3.  The  dividend  is  equal  to  the  divisor  multiplied  by  the 
quotient.     Let  q  =  9,  m  =  99,  n  =  11  ;  q  =  6,  m  =  42,  n  =  7. 

4.  The  divisor  is  equal  to  the  dividend  divided  by  the  quo- 
tient.    Let  q  =  5,  m  =  45,  n  =  9  ;    q  =  6,  m  =  72,  n  =  12. 

State  in  verbal  language  the  principles  which  are  expressed 
in  symbols  in  the  following  : 

6    a     c  _  a  x  c 
b     d     b  x  d 


a  _  a 


d     d  x  n  d     d  -*-  n 

11-16.    In  Exs.  5-10,  let  a  =  8,  6  =  7,  c  =  5,  d  =  6,  ?i  =  2  ; 
a  =  15,  6  =  8,  c  =  11,  d  —  5,  n  =  5. 


3-8]  GENERAL   NUMBER.  S 

5.  As  was  assumed  in  Art.  2,  the  operations  of  Addition, 
Subtraction,  Multiplication,  and  Division  are  denoted  by  the 
same  symbols  in  Algebra  as  in  Arithmetic. 

Just  as  5  +  3,  read  five  plus  three,  means  that  3  is  to  be 
added  to  5 ;  so  a  +  b,  read  a  plus  b,  means  that  b  is  to  be 
added  to  a. 

Just  as  5  —  3,  read  Jive  minus  three,  means  that  3  is  to  be 
subtracted  from  5 ;  so  a  —  b,  read  a  minus  b,  means  that  b  is 
to  be  subtracted  from  a. 

Just  as  5  x  3,  read  Jive  multiplied  by  three,  means  that  5  is 
to  be  multiplied  by  3 ;  so  a  x  b,  read  a  multiplied  by  b,  means 
a  is  to  be  multiplied  by  b. 

Just  as  10  -T-  5,  read  ten  divided  by  Jive,  means  that  10  is  to 
be  divided  by  5;  so  a-t-  b,  read  a  divided  by  b,  means  that  a  is 
to  be  divided  by  b. 

6.  Since  5x3  =  3x5,  either  number  may  be  taken  as  the 
multiplier,  the  other  as  the  multiplicand. 

If  the  number  on  the  left  be  taken  as  the  multiplier,  the 
symbol  of  multiplication  is  read  times  or  into. 

As,  5x3,  read  five  times  three,  if  5  be  regarded  as  the 
multiplier. 

A  dot  (•)  is  frequently  used,  instead  of  the  symbol  x,  to 
denote  multiplication ;  as  a  •  b  for  a  x  b. 

7.  The  symbol  of  multiplication  between  two  literal  num- 
bers, or  one  literal  number  and  an  Arabic  numeral,  is  fre- 
quently omitted. 

E.g.,  the  product  x  x  y  X  z,  or  x  •  y  •  z,  is  usually  written  xyz, 
and  is  read  x-y-z. 

But  the  symbol  of  multiplication  between  two  numerals 
cannot  be  omitted  without  changing  the  meaning. 

E.g.,  if  in  the  indicated  multiplication,  3x6,  or  3-6,  the 
symbol,  x ,  or  •,  were  omitted,  we  should  have  36,  not  18. 

8.  In  a  chain  of  additions  and  subtractions  the  operations 
are  to  be  performed  successively  from  left  to  right. 

E.g.,        7  +  4-3  +  2  =  11 -3  +  2  =  8  +  2  =  10. 


4  ALGEBRA.  [Cn.  I 

In  a  chain  of  multiplications  and  divisions  the  operations 
are  to  be  performed  successively  from  left  to  right. 

E.g.,      12  x  2  -^  3  x  4  =  24  -=-3x4  =  8x4  =  32. 

In  a  chain  of  additions,  subtractions,  multiplications,  and 
divisions,  the  multiplications  and  divisions  are  first  to  be  per- 
formed, and  then  the  additions  and  subtractions. 

Kg.,  2x3  +  8-v-4  =  6  +  2  =  8. 

9.  An  Algebraic  ^Expression  is  a  number  expressed  by  means 
of  the  signs  and  symbols  of  Algebra ;  as  x,  mn,  ab  —  cd,  etc. 

10.  The  Symbol  of  Equality,  =,  read  is  equal  to,  is  placed 
between  two  numbers  to  indicate  that  they  have  the  same  or 
equal  values  ;  as  3  -f-  2  =  5. 

11.  The  Symbol  of  Inequality,   >,  read   is  greater  than,  is 
used  to  indicate  that  the  number  on  its  left  is  greater  than 
that  on  its  right ;  as  7  >  5. 

12.  The  Symbol  of  Inequality,  <,  read  is  less  than,  is  used  to 
indicate  that  the  number  on  its  left  is  less  than  that  on  its 
right ;  as  3  <  4  +  2. 

13.  The  use  of  letters  to  represent  general  numbers  may  be 
illustrated  by  a  few  simple  examples. 

Ex.  1.  If  a  boy  has  3  books  and  is  given  2  more,  he  will 
have  3  +  2  books.  If  he  has  a  books  and  is  given  5  more,  he 
will  have  a  -+-  5  books.  If  he  has  m  books  and  is  given  n  more, 
he  will  have  m  +  n  books. 

Ex.  2.  If  a  man  buys  5  city  lots  at  120  dollars  each,  he  pays 
120  x  5  dollars  for  the  lots.  If  he  buys  a  lots  at  150  dollars 
each,  he  pays  150  a  dollars  for  the  lots.  If  he  buys  u  lots  at  v 
dollars  each,  he  pays  vu  dollars  for  the  lots. 

Ex.  3.  If  a  train  runs  60  miles  in  two  hours,  it  runs  60  -=-  2 
miles  in  1  hour.  If  it  runs  a  miles  in  5  hours,  it  runs  a  -4- 5 
miles  in  1  hour.  If  it  runs  p  miles  in  q  hours,  it  runs  p  -*-  q 
miles  in  1  hour. 


8-13]  GENERAL  NUMBER.  5 

Ex.  4.  If,  in  a  number  of  two  digits,  the  digit  in  the  units? 
place  is  3  and  the  digit  in  the  tens'  place  is  5,  the  number  is 
10  x  5  +  3.  If  the  digit  in  the  units'  place  is  a  and  the  digit 
in  the  tens'  place  is  6,  the  number  is  10  b  +  a. 

Ex.  5.    Just  as  2  =  1  +  1,  and     3  =  1+1  +  1, 
so  2  a  =  a  +  a,  and  3a  =  a  +  a  +  a. 

Therefore,  just  as  3  +  2  —  5,  so  3  a  +  2  a  =  5  a. 

In  like  manner,          5  x  —  3  x  =  2  x^ 
and  $x  +  %x  =  %x. 

EXERCISES   II. 

Read  the  following  expressions  : 

1.    a  +  b.  2.   ra  —  n.  3.    a  x  b.  4.   m  -t- n. 

5.   ±x  +  2y.      6.   3m-Sn.      7.   4  a  x  5  6.  8.    7  aj  -J-  3  y. 

9.    a  +  6  +  c.  10.    a?  —  y  +  2.  11.  m  —  n  —  p. 

12.   4  a  —  c  +  3  d.         13.    a&  +  be  —  ac.       14.  3  xy  —  5  6cd. 

15.  A  father  is  n  years  older  than  his  son.     How  old  is  the 
father,  if  the  son  is  10  years  old  ?     If  the  son  is  x  years  old  ? 

16.  A  boy  rides  his  bicycle  x  miles  and  then  walks  y  miles. 
How  many  miles  does  he  go  altogether  ? 

17.  A  man  has  $  d.     If  he  spends  $  10,  how  many  dollars 
has  he  left  ?     If  he  spends  $  z,  how  many  dollars  has  he  left  ? 

18.  A  man  is  now  n  years  old.    How  old  was  he  8  years  ago  ? 
m  years  ago?     How  long  must  he  live  to  be  75  years  old? 
How  long  to  be  y  years  old  ? 

19.  If  one  pencil  costs  3  cents,  how  much  do  5  pencils  cost? 
x  pencils  ? 

20.  If  one  pencil  costs  c  cents,  how  much  do  z  pencils  cost  ? 

21.  How  many  square  feet  are  there  in  a  floor  15  feet  long 
and  20  feet  wide  ?     In  a  floor  a  feet  long  and  b  f eet  wide  ? 


6  ALGEBRA.  [Cn.  1 

22.  A  train  runs  m  miles  in  1  hour.     How  many  miles  will 
it  run  in  4  hours  ?     In  b  hours  ? 

23.  A  train  runs  m  miles  in  8  hours.     How  many  miles  will 
it  run  in  1  hour?     If  it  runs  m  miles  in  h  hours,  how  many 
miles  will  it  run  in  1  hour  ? 

24.  A  boy  paid  c  cents  for  5  note-books.     How  much  did  he 
pay  for  each  ?     If  he  paid  c  cents  for  n  note-books,  how  much 
did  he  pay  for  each  ? 

25.  In  3  dimes  there  are  10  x  3  cents.     How  many  cents  in 
d  dimes  ?     In  x  dimes  ? 

26.  How  many  cents  in  a  dollars  and  b  dimes  ?    In  x  dollars, 
y  dimes,  and  z  cents  ? 

27.  10  x  2,  10  x  3,  10  x  4,  etc.,  are  particular  multiples  of 
10.     Write  any  multiple  of  10. 

28.  Write  a  number  containing  8  units  and  5  tens.    Contain- 
ing u  units  and  t  tens. 

29.  Write  a  number  containing  h  hundreds,  t  tens,  and  u 
units.     Containing  a  hundreds,  b  tens,  and  c  units. 

WThat  are  the  values  of  the  following  expressions  ? 

30.  a  +  a.  31.  a  +  2 a.  32.    x  +  3x. 
33.  a  —  a.  34.  2  a  —  a  35.    3z  — z. 
36.  3  c  +  5  c.  37.  5  d  —  3  d.  38.  8  x  +  5  x. 
39.  Sx  —  5x.  40.  x  +  %x.  41.    x  —  ^x. 
42.  f  a  4-  i  a.  43.  |  a  —  i  a.  44.    5  m  —  f  m. 
45.  a  4  2 a  +  3 a.  46.  a  +  2 a  —  3 a.  47.    5z  +  8z  +  4z. 
48.  8z  —  52-f43.  49.  9x4- 3 x  — 8 x.  50.   9#  —  4?/  —  :$.". 

51.  A  man  has  $10#.     If  he  receives  $8x,  how  many  dol- 
lars will  he  have?     If  he  spends  $6x,  how  many  dollars  will 
he  have  left  ? 

52.  A  boy  paid  Sx  cents  for  pencils  and  Sx  cents  for  note- 
books.   How  much  did  he  pay  for  both  ?    How  much  more  for 
note-books  than  for  pencils? 


13-15]  GENERAL   NUMBER.  7 

53.  A  girl  has  x  dimes  and  3  x  cents.     How  many  cents  has 
she? 

54.  A  girl  has  a  dollars.    If  she  spends  7  a  dimes,  how  many 
dimes  will  she  have  left?     If  she  spends  85 a  cents,  how  many 
cents  will  she  have  left  ? 

55.  A  man  has  $  45  x.    If  he  spends  $  7  x  for  a  lot,  and  $  32  x 
for  a  house,  how  many  dollars  will  he  have  left? 

56.  A  boy  rides  a  wheel  x  miles  and  then  walks  160  x  rods. 
How  many  rods  did  he  go  altogether  ?     How  many  rods  more 
did  he  ride  than  walk  ? 

57.  The  width  of  a  room  is  x  yards,  and  the  length  is  2  x 
feet  greater  than  the  width.     How  many  feet  are  there  in  the 
length  of  the  room? 

Axioms. 

14.  An  Axiom  is  a  truth  so  simple  that  it  cannot  be  made 
to  depend  upon  a  truth  still  simpler. 

Algebra  makes  use  of  the  following  mathematical  axioms : 
(i.)  Every  number  is  equal  to  itself.     E.g.,  1  =  1,  a  =  a. 
(ii.)  The  whole  is  equal  to  the  sum  of  all  its  parts. 
E.g.,  7  =  3  +  4,  5  =  1  +  1+1  +  1+1. 

(iii.)  If  two  numbers  be  equal,  either  can  replace  the  other  in 
any  algebraic  expression  in  ivhich  it  occurs. 

E.g.,  If  a-\-b  =  c,  and  b  =  2,  then  a+2  =  c,  replacing  b  by  2. 

(iv.)   Two  numbers  which  are  each  equal  to  a  third  number  are 
equal  to  each  other. 

E.g.,  If  a  =  b,  and  c  =  b,  then  a  =  c. 

(v.)   The  ivhole  is  greater  than  any  of  its  parts;   and,  con- 
versely, any  part  is  less  than  the  whole. 
E.g.,  3  +  2  >  2  and  2  <  3  +  2. 

15.  Literal  numbers,  as  has  been  stated,  are  numbers  which 
may  have  any  values  whatever.     But  it  is  frequently  necessary 
to  assign  particular  values  to  such  numbers. 


8  ALGEBRA.  [Cn.  I 

16.  Substitution  is  the  process  of  replacing  a  literal  number 
in  an  algebraic  expression  by  a  particular  value.     See  axiom 
(iii.).  Simple   examples   in   substitution  have   already  been 
given  in  Art.  2. 

Ex.  1.    If,  in  a  +  b,  we  let  a  =  3  and  b  =  5,  then 
a  +  6  =  3  +  5  =  S,  or  a  +  6  =  S. 

Ex.  2.    If,  in   a  +  b  —  2  a  +  3  b  —  c,   we   let   a  =  6,    6  =  11, 
c  =  1,  we  have 

a  +  6-2a  +  36-c  =  6  +  ll-2x6  +  3xll-l 
=  6  +  11-12  +  33-1-37. 

Ex.  3.    If,  in  the  last  example,  a  —  3,  6  =  1,  and  c  =  1,  we 
have  a  +  6-2a  +  36-c  =  3  +  l-6  +  3-l=4-6  +  3-l. 

We  cannot  further  reduce  4—6+3  —  1,  since  we  are  unable, 
as  yet,  to  subtract  6  from  4. 

EXERCISES   III. 

When  a  =  10,  6  =  5,  c  =  3,  find  the  values  of  the  following 
expressions : 

1.  a  +  6.  2.  a  —  6.  3.  a6. 

4.  a  -r-  6.  5.  a  +  6  —  c.  6.  a  —  6  +  c. 

7.  a  _  &  _  c.  8.  c  +  3  a.  9.  5  6  —  3  c. 

10.  2a  +  36-5c.  11.  5a-26-6c.  12.  3a-56  +  8c. 

13.  lab.  14.  2a6c.  15.  3a66. 

16.  2  a6  +  5  ac.  17.  3  ac  —  5  6c.  18.  5  aa  —  3  66. 

19.   2  a6  —  3  ac  +  5  6c.                     20.  5  aa  —  3  66  +  6  cc. 

Fundamental  Principles. 

17.  The  following  principles  are  obtained  directly  from  the 
axioms : 

(i.)  If  the  same  number,  or  equal  numbers,  be  added  to  equaf 
numbers,  the  sums  will  be  equal. 


16-19]  GENERAL   NUMBER.  9 

(ii.)  If  the  same  number,  or  equal  numbers,  be  subtracted  from 
equal  numbers,  the  remainders  will  be  equal. 

(iii.)  If  equal  numbers  be  multiplied  by  the  same  number,  or 
by  equal  numbers,  the  products  will  be  equal. 

(iv.)  If  equal  numbers  be  divided  by  the  same  number  (except  0), 
or  by  equal  numbers,  the  quotients  will  be  equal. 

E.g.,  if  3  x  =  6, 

then  3^      2  =  6      2      3x-5  =  6-5, 


Equations. 

18.  An  Equation  is  a  statement  that  two  expressions  are 
equal  ;  as  7  x  9  =  63,  4  x  7  +  3  =  31. 

The  First  Member  of  an  equation  is  the  expression  on  the  left 
of  the  symbol  =  ;  the  Second  Member  is  the  expression  on  the 
right  of  the  symbol  =. 

19,  Ex.  1.    What  is  the  value  of  x  in  the  equation 

3  x  +  8  x  =  22  ? 
Since  3  x  +  8  x  =  11  x,  we  have 

11  a?  =  22. 
Dividing  both  members  by  11  [Art.  17,  (iv.)], 

x  =  2. 

To  check  this  result  we  substitute  2  for  x  in  the  equation 
3a  +  8z  =  3x2+8.x2  =  6-fl6  =  22. 

Ex.  2.    If  8  x  —  3  x  has  the  value  20,  what  is  the  value  of  x  ? 
We  have  8  x  -  3  x  =  20. 

Or,  since  Sx  —  3x  =  5x,          5  a?  =  20. 
Dividing  both  members  by  5,    x  —  4. 
Check:    8x4-3x4-32-12  =  20. 


10  ALGEBRA.  [Cii.  I 

20.  An  Unknown  Number  of  an  equation  is  a  number  whose 
value  is  to  be  found  from  the  equation. 

The  Known  Numbers  of  an  equation  are  the  numbers  whose 
values  are  given. 

In  the  equation  x  -f  1  =  3, 

the  unknown  number  is  a?,  and  the  known  numbers  are  1  and  3. 

Unknown   numbers   are   usually   represented   by   the   final 

letters  of  the  alphabet,  x,  y,  z,  etc.,  as  in  the  above  examples. 

EXERCISES   IV. 

Find  the  value  of  x  in  each  of  the  following  equations : 

1.  3x  =  9.  2.  6 x  =  18.  3.   5x  =  0. 

4.  ix  =  4.  5.  ix  =  5.  6.   ia  =  0. 

7.  fo;  =  6.  8.  |a?  =  15.  9.    £a  =  21. 

10.  x  +  x  =  &.  11.  a?  +  5  a;  =  24.        12.    5  #  +  4  ;E  =  45. 

13.  5a-4#  =  3.       14.  6x  —  3x  =  3.       15.    7a;-5z=12. 

16.  x  4-  3 x  +  5 a;  =  18.  17.    2oj  +  5a;  +  3a;  =  20. 

18.  7a  +  3oj  +  5a;  =  90.  19.    5  .T  +  4  a?  -  6  x  =  15. 

20.  8  a;  —  5  x  +  x  =  12.  21.    11  a;  +  7  a;  —  5  a;  =  26. 
22.  x  +  £  x  =  6.                             23.    a?  —  i  a  =  10. 

24.  24^  +  |«  =  149.  25.  3a;  +  f»  =  30. 

26.  5x-%x  =  33.  27.  2%x-±x  =  U. 

28.  x  +  ^x  +  f^  =  28.  29.  2.T-^x  +  |^  =  34. 

30.  Jaj  +  f  a;-ia;  =  54.  31.  5  x  -  \x  -  ^x  =  62. 

Problems  solved  by  Equations. 

21.  A  Problem  is  a  question  proposed  for  solution. 
Another  use  of  literal  numbers  is  shown  by  the  following 

problems : 

Pr.  l.  The  older  of  two  brothers  has  twice  as  many  marbles 
as  the  younger,  and  together  they  have  33  marbles.  How 
many  has  the  younger? 


20-21]  GENERAL  NUMBER.  11 

The  number  of  marbles  the  younger  brother  has  is,  as  yet, 
an  unknoivn  number. 

Let  us  represent  this  unknown  number  by  some  letter,  say  x. 

Then,  since  the  older  brother  has  twice  as  many,  he  has 
2  x  marbles. 

The  problem  states, 

in  verbal  language  :  the  number  of  marbles  the  younger  has  plus 
the  number  the  older  has  is  equal  to  33 ; 

in  algebraic  language,    x  +  2  x  =  33, 
or,  3  x  =  33. 

Dividing  both  members  of  the  last  equation  by  3,  we  have 

x  =  n, 

the  number  of  marbles  the  younger  has. 

The  older  has,     2  x,  =  2  x  11,  =  22  marbles. 

To  check  this  result,  we  substitute  11  for  x  in  the  equation 
of  the  problem : 

x  +  2x=ll+22  =  33. 

Notice  that  the  letter  x  stands  for  an  abstract  number.  The 
beginner  must  never  put  x  for  marbles,  distance,  time,  etc., 
but  for  the  number  of  marbles,  of  miles,  of  hours,  etc. 

Pr.  2.  Divide  52  into  three  parts,  so  that  the  second  shall 
be  one-half  of  the  first,  and  the  third  one-fourth  of  the  second. 

Let  x  stand  for  the  first  part. 

Then  |  x  stands  for  the  second  part, 

and  J  x  \  x,   =  J  x,  stands  for  the  third  part. 

The  problem  states, 

in  verbal  language :  the  first  part,  plus  the  second  part,  plus  the 
third  part,  is  equal  to  52  ; 

in  algebraic  language,   x  +  \  x  -f-  \  x  —  52, 
or,  J£  x  =  52. 


12  ALGEBRA.  [Cn  I 

Dividing  both  members  of  the  last  equation  by  13, 

*•-*• 

Multiplying  both  members  of  this  equation  by  8, 

x  =  32, 
the  first  part.     Then  the  second  part  is 

1  x,  =  i  x  32,   =  16, 
and  the  third  part  is 

-L  x,  =  i  X  32,  =4. 

Check:          x  +  ±  x  +  ±x  =  32 +  16  +  4:  =  52. 

22.  In  stating  problems  in  algebraic  language,  the  beginner 
should  observe  the  following  directions : 

(i.)  Read  the  problem  carefully,  and  note  what  are  the  numbers 
whose  values  are  required. 

(ii.)  Let  some  letter,  say  x,  stand  for  one  of  the  required 
numbers. 

(iii.)  The  problem  will  contain  statements  about  the  values'  of 
other  numbers.  Use  these  statements  to  express  their  values  in 
terms  of  x. 

(iv.)  Express  concisely  in  verbal  language  a  statement  in  the 
problem  which  furnishes  an  equation. 

(v.)  Express  this  statement  in  algebraic  language. 

EXERCISES   V. 

1.  What  number  is  five  times  x?     Twelve  times  x? 

2.  Five  times  a  number  is  80.     What  is  the  number  ? 

3.  Twelve  times  a  number  is  132.     What  is  the  number  ? 

4.  The  greater  of  two  numbers  is  four  times  the  less.     If 
the  less  is  x,  what  is  the  greater  ?     What  is  their  sum  ?     Their 
difference  ? 

5.  The  greater  of  two  numbers  is  four  times  the  less.     If 
their  sum  is  75,  what  are  the  numbers  ? 


21-22]  GENERAL  NUMBER.  13 

6.  The  greater  of  two  numbers  is  seven  times  the  less.     If 
their  difference  is  72,  "what  are  the  numbers  ? 

7.  A  father  is  three  times  as  old  as  his  son.     If  the  son  is 
x  years  old,  how  old  is  the  father  ?     What  is  the  sum  of  their 
ages  ?     How  much  older  is  the  father  than  the  son  ? 

8.  A  father  is  three  times  as  old  as  his  son,  and  the  sum  of 
their  ages  is  48  years.     How  old  is  each  ? 

9.  A  father  is  five  times  as  old  as  his  son.     If  the  father  is 
32  years  older  than  his  son,  what  are  their  ages  ? 

10.  At  an  election  A  received  twice  as  many  votes  as  B,  and 
his  majority  was  J.38.     How  many  votes  did  each  receive  ? 

11.  In  a  company  are  32  persons.     The  number  of  children 
is  three  times  the  number  of  adults.     How  many  are  there  of 
each? 

12.  Two  trains   leave  Philadelphia  in  opposite  directions. 
After  one  hour  they  are  60  miles  apart.     If  one  has  gone  three 
times  as  far  as  the  other,  how  many  miles  is  each  from  Phila- 
delphia ? 

13.  Two  trains  leave  Chicago  in  the  same  direction.     After 
one  hour  they  are  20  miles  apart.     If  one  has  gone  twice  as  far 
as  the  other,  how  far  is  each  from  Chicago  ? 

14.  A  man  pays  $  55  in  one-dollar  bills  and  ten-dollar  bills. 
If  he  pays  the  same  number  of  one-dollar  bills  as  of  ten-dollar 
bills,  how  many  of  each  does  he  pay  ? 

15.  In  a  number  of  two  digits,  the  tens'  digit  is  three  times 
the  units'   digit,  and  their  sum  is  8.     What  are  the  digits  ? 
W^hat  is  the  number  ? 

16.  In  a  number  of  two  digits,  the  units'  digit  is  twice  the 
tens'  digit,  and  their  difference  is  3.     What  is  the  number  ? 

17.  What  is  the  sum  of  twice  x  and  six  times  x  ?     The 
difference  ? 

18.  If  twice  a  number  is  added  to  six  times  the  same  num- 
ber, the  sum  will  be  96.     What  is  the  number  ? 


14  ALGEBRA.  [Cn.  I 

19.  If  four  times  a  number  is  subtracted  from  seven  times 
the   same  number,  the  remainder   will   be  72.     What  is  the 
number  ? 

20.  A  traveller  first  rides  his  bicycle  9  miles  an  hour.     He 
then  rides  the  same  number  of  hours  in  a  car  35  miles  an  hour. 
If  he  travels  132  miles,  how  many  hours  did  he  ride  his  bicycle  ? 

21.  Two  trains  run  out  of  New  York  in  opposite  directions. 
One  runs  42  miles  an  hour,  the  other  34  miles  an  hour.     After 
how  many  hours  will  they  be  228  miles  apart  ? 

22.  Two  trains  run  out  of  New  York  in  the  same  direction. 
One  runs  38  miles  an  hour,  the  other  34  miles  an  hour.     After 
how  many  hours  will  they  be  32  miles  apart  ? 

23.  A  boy  has  75  cents  in  dimes  and  five-cent  pieces.     He 
has  the  same  number  of  dimes  as  of  five-cent  pieces.     How 
many  coins  of  each  kind  has  he  ? 

24.  A  owes  B  $  40.     He  pays  his  debt  in  ten-dollar  bills, 
and  receives  in  change  the  same  number  of  two-dollar  bills. 
How  many  ten-dollar  bills  did  A  pay  B  ? 

25.  A  cistern  has  two  pipes.     One  lets  in  8  gallons  a  minute, 
and  the  other  15  gallons  a  minute.     If  the  cistern  holds  207 
gallons,  how  many  minutes  will  it  take  the  pipes  to  fill  it  ? 

26.  A  cistern  has  two  pipes.     One  lets  in  11  gallons  a  minute, 
and  the  other  lets  out  6  gallons  a  minute.     How  many  minutes 
will  it  take  the  one  pipe  to  let  in  85  gallons  more  than  the 
other  lets  out  ? 

27.  What  is  the  sum  of  x,  four  times  x,  and  seven  times  x? 
Of  oj,  twice  x,  and  five  times  x  ? 

28.  The  sum  of  a  certain  number,  four  times  the  number, 
and  seven  times  the  number  is  96.     What  is  the  number  ? 

29.  Three  boys,  A,  B,  and  C,  together  have  21  pencils.     B 
has  twice  as  many  as  A,  and  C   four  times  as  many  as  A. 
How  many  has  A  ?     How  many  has  each  ? 

30.  Divide  147  into  three  parts,  so  that  the  second  part  shall 
be  four  times  the  first,  and  the  third  part  twice  the  first. 


22]  GENERAL   NUMBER.  15 

31.  A  merchant  receives  $  64  in  ten-dollar  bills,  five-dollar 
bills,  and  one-dollar  bills.     He  receives  the  same  number  of 
each  kind.     How  many  of  each  does  he  receive  ? 

32.  At  an  election  726  votes  were  cast.     A,  B,  and  C  were 
candidates.     B  received  three  times  as  many  votes  as  C,  and  A 
twice  as  many  as  C.     How  many  votes  did  each  receive  ? 

33.  A  cistern  has  three  pipes.     The  first  lets  in  6  gallons  a 
minute,  the  second  9  gallons  a  minute,  and  the  third  12  gallons 
a  minute.     If  the  cistern  holds  243  gallons,  how  long  will  it 
take  the  pipes  to  fill  it  ? 

34.  A  cistern  has  three  pipes.     The  first  lets  in  5  gallons  a 
minute,  the  second  14  gallons  a  minute,  and  the  third  lets  out 
10  gallons  a  minute.     How  many  minutes  will  it  take  the  two 
pipes  to  let  in  108  gallons  more  than  the  third  pipe  lets  out  ? 

35.  An  estate  of  $  9600  is  divided  among  2  sons  and  2 
daughters.     The  sons  receive  equal  amounts,  and  a  daughter 
receives  three  times  as  much  as  a  son.     How  many  dollars 
does  each  receive  ? 

36.  What  is  twice  3  x  ?     Seven  times  5  x  ?     Four  times  9  x  ? 

37.  A  receives  x  dollars,  B  receives  three  times  as  much  as 
A,  and  C   receives  twice  as  much  as  B.     How  many  dollars 
does  C  receive  ?     How  many  dollars  do  all  receive  ? 

38.  Three  boys,  A,  B,  and  C,  together  receive  $  70.     B  re- 
ceives three  times  as  much  as  A,  and  C  twice  as  much  as  B. 
How  many  dollars  does  each  receive  ? 

39.  A  merchant's  profits  doubled  each  year  for  three  years. 
If  his  profits  for  the  three  years  were  $  8750,  what  were  his 
profits  the  first  year  ? 

40.  In  a  company  are  50  persons.     The  number  of  women  is 
three  times  the  number  of  men,  and  the  number  of  children  is 
twice  the  number  of  women.     How  many  of  each  are  in  the 
company  ? 

41.  What  number  is  \  of  a?  ?     f  of  x  ? 


16  ALGEBRA.  [Cn.  I 

42.  If  J  of  a  number  is  16,  what  is  the  number  ? 

43.  The  less  of  two  numbers  is  f  of  the  greater.     If  the 
greater  is  x,  what  is  the  less  ?     What  is  their  sum  ?     Their 
difference  ? 

44.  The  less  of  two  numbers  is  f  of  the  greater.     If  their 
sum  is  91,  what  are  the  numbers  ? 

45.  A  and  B  together  have  $  1133.     If  B  has  -^  as  much  as  A, 
how  many  dollars  has  each  ? 

46.  A  has  $  31  more  than  B.     If  B  has  J  as  much  as  A,  how 
many  dollars  has  each  ? 

47.  Two  boys,  A  and  B,  catch  36  fish.     If  A  catches  f  as 
many  as  B,  how  many  fish  does  each  catch  ? 

48.  A  workman  pays  f  of  his  wages  for  board.     If  he  has 
left  $  8  each  week,  what  are  his  wages  ? 

49.  Two    boys   together   solve   65   problems.     If   the  first 
solves   -f  as  many  as   the  second,  how  many  problems  does 
each  solve  ? 

50.  A  solves  21  more  problems  than  B.     If  B  solves  J-  as 
many  as  A,  how  many  problems  does  each  solve  ? 

51.  A  tree  126  feet  high  is  broken  by  the  wind.     If  the  part 
left  standing  is  T3T  of  the  part  broken  off,  how  long  is  each 
part  ? 

52.  What  is  the  sum  of  -J-  of  x  and  f  of  x  ?     The  difference  ? 

53.  If  i  of  a  number  is  added  to  J  of  the  same  number,  the 
sum  will  be  39.     What  is  the  number  ? 

54.  If  -|  of  a  number  is  subtracted  from  J  of  the  same  num- 
ber, the  remainder  will  be  3.     What  is  the  number  ? 

55.  If  to  a  number  is  added  -J-  of  itself  and  f  of  itself,  the 
sum  will  be  50.     What  is  the  number  ? 

56.  Three  boys,  A,  B,  and  C,  together  have  29  pencils.     B 
has  |  as  many  as  A,  and  C  has  }  as  many  as  A.     How  many 
pencils  has  each  ? 


22-23]  GENERAL   NUMBER.  17 

57.  Divide  104  into  three  parts,  so  that  the  first  shall  be 
three  times  the  second,  and  the  third  ^  of  the  second. 

58.  A  man  makes  a  journey  of  69  miles.     He  goes  f  as  far 
by  boat  as  by  train,  and  ±  as  far  by  stage  as  by  train.     How 
many  miles  does  he  go  by  each  conveyance  ? 

59.  What  is  £  of  three  times  x?     Twicefofar?     f  off  of  a? 

60.  The  second  of  three  numbers  is  three  times  the  first,  and 
the  third  is  -i-  of  the  second.     If  the  first  number  is  x,  what 
is  the  second  ?     The  third  ?     What  is  the  sum  of  the  three 
numbers  ?  * 

61.  The  sum  of  three  numbers  is  99.     The  second  is  four 
times  the  first,  and  the  third  is  f  of  the  second.     What  are 
the  numbers  ? 

62.  The  width  of  a  field  is  \  of  its  length,  and  the  distance 
around  it  is  88  rods.     What  is  the  width  and  the  length  of 
the  field  ? 

63.  The   sum  of   $  420   is   divided   among   A,    B,    and   C. 
B   receives   f   as   much  as  A,  and  C   as  much  as  A  and   B 
together.     How  many  dollars  does  each  receive  ? 

64.  A  sells  a  number  of  apples  at  2  cents  apiece,  and  B  sells 
\  as  many  at  3  cents  apiece.     If  they  receive  together  87  cents, 
how  many  apples  does  each  sell  ? 

Parentheses. 

23.   Parentheses,  (),  and  Brackets,  [],  are  used  to  indicate 
that  whatever  is  placed  within  them  is  to  be  treated  as  a  whole. 

E.g.,  10  —  (2  +  5)  means  that  the  result  of  adding  5  to  2,  or 
7,  is  to  be  subtracted  from  10 ;  that  is, 

10  -  (2  +  5)  =  10  -  7  =  3. 

But  10  —  2  +  5  means  that  2  is  to  be  subtracted  from  10  and 
5  is  then  to  be  added  to  that  result ;  that  is, 

10  _  2  +  5  =  8  +  5  =  13. 


18  ALGEBRA.  [Cn.  I 

In  like  manner,  [27  —  (3  -f  2)  x  5]  -H  2  means  that  the  result 
of  multiplying  the  sum  3  -f  2  by  5  is  first  to  be  subtracted 
from  27,  and  the  remainder  is  then  to  be  divided  by  2  ;  that  is, 

[27  -  (3  +  2)  x  5]  -*•  2  =  [27  -  25]  •+•  2  =  2  -*-  2  =  1. 
Likewise,  (a+fr)c  is  the  result  of  multiplying  a+b  by  c,  etc. 


EXERCISES   VI. 

Find  the  value  of  each  of  the  following  expressions  : 

1.  10  +  (3  +  2).        2.   10  -(3  +  2).  3.   10  +  (3  -2). 

4.  10  -(3  -2).        5.   27  -(18  -11).       6.   53  +  (40  +  7). 

7.  97  -f  (11  -8).      8.   58  -(15  -7).         9.   99  +  (18-  17). 

10.  5(8  +  2).  11.   6(11-6).  12.    (10  +  15)-*-  5. 

13.  10  +  (15-s-5).      14.    (12  -4)  -4-  2.         15.    12  -(4-5-2). 

16.  (15  -  3)  +  (18  -  6).  17.    (16  -  2)  -  (20  -  8). 

18.  (4  +  5)  (8  -3).  19.    (8  +  12)  -  (7  -  2). 

20.  20  +  [11  -  (5  +  2)].  21.   28  -[16  -(5  +  3)]. 

22.  [26  -  (14  +  6)]  x  5.  23.    [27  -  (18  -  12)]  -j-  7. 

When  a  =  12,  b  =  6,  c  =  3,  find  the  values  of  : 
24.    a  +  (b  —  c).  25.  a  —  (6  +  c).  26.  a  —  (b  —  c). 

27.   c  +  5(a-6).        28.  4a-2(6  +  c).     29.  6[c  +  (a-6)]. 
30.   a[a-~K&+c)].     31.   [a-(6-c)]-s-c.    32.  [&  +  (a-c)]-s-c. 


POSITIVE   AND   NEGATIVE   NUMBERS,   OR  ALGEBRAIC 
NUMBERS. 

24.  In  ordinary  Arithmetic  we  subtract  a  number  from  an 
equal  or  a  greater  number.  We  are  familiar  with  such  opera- 
tions as 

543          minuend 
333         subtrahend  (i.) 

210          remainder 


23-26]         POSITIVE   AND  NEGATIVE  NUMBERS.  19 

But  such  operations  as 

210  minuend 

333  subtrahend  (ii.) 

?  ?  ?  remainder 

have  not  occurred  in  ordinary  Arithmetic.  In  Arithmetic  we 
cannot  subtract  from  a  number  more  units  than  are  contained 
in  the  number. 

25.  Now,  in  the  operations  (i.)  above,  the  remainder  2  indi- 
cates that  the  subtrahend  is  two  units  less  than  the  minuend ; 
the  remainder  1  that  the  subtrahend  is  one  unit  less  than  the 
minuend ;  and  the  remainder  0  that  the  subtrahend  is  equal  to 
the  minuend. 

In  the  operations  (ii.),  we  must  indicate  by  the  remainders 
that  the  subtrahend  is  one,  tivo,  three,  etc.,  units  greater  than 
the  corresponding  minuend. 

We  do  this  by  placing  the  sign  ~  before  the  symbols  for  one, 
two,  three,  etc. ;  as  ~1,  ~2,  ~3,  etc. 

The  remainders  in  these  cases  are  called  Negative  Numbers ; 
as  ~1,  ~2,  ~3,  etc.,  read  negative  one,  negative  two,  negative 
three,  etc. 

For  the  sake  of  distinction,  the  remainders  in  the  operations 
(i.)  are  called  Positive  Numbers. 

They  are  indicated  by  the  sign  + ;  as  +1,  +2,  +3,  etc.,  read 
positive  one,  positive  two,  positive  three,  etc. 

Positive  and  negative  numbers  are  called  Algebraic  or  Rela- 
tive Numbers. 


26.  We 

pos.  5 
pos.  3 

can  now 

pos.  4 
pos.  3 

write  (i 

pos.  3 
pos.  3 

)  and 

pos. 
pos. 

(ii.)  as  follows 

2    pos.  1 
3    pos.  3    pos. 

0 
3 

inin. 
sub. 

rem. 

min. 
sub. 

rem. 

pos.  2 

+5 
+3 

+2 

pos.  1 

+4 
+3 

+1 

0 

+3 
+3 

0 

neg.l 

+2 
+3 

neg.2 

+1 
+3 

-2 

neg. 

-f 

3 

0 

3 

(iii.) 


20  ALGEBRA.  [Cn.  I 

27.  We  thus  have  in  Algebra  the  series  of  numbers, 

.   +*»  +4  +3  +<>  +1    o  -1    -9  -3  -A  -*> 

>       °|       *J       °J       -'J       •••>   UJ       -1-?       "J       °)       ^7       O,    •", 

wherein  the  signs,  •••,  indicate  that  the  succession  of  numbers 
continues  without  end  in  both  directions.  This  series  is  usu- 
ally written  with  the  positive  numbers  on  the  right,  as 

...,  -5,  -4,  -3,  -2,  -1,  0,  +1,  +2,  +3,  +4,  +5,  .... 

28.  In  this  series  the  numbers  increase  by  one  from  left  to 
right,  and  decrease  by  one  from  right  to  left.     Or,  a  number  is 
greater  than  any  number  on  its  left,  and  less  than  any  number 
on  its  right. 

Thus,  +2  is  one  unit  greater  than  +1,  two  units  greater  than 
0,  three  units  greater  than  ~1,  etc.  Again,  ~3  is  three  units 
greater  than  ~6,  two  units  less  than  ~1,  three  units  less  than 
0,  etc. 

29.  The  signs  +  and  ~  are  called  signs  of  quality;  the  signs 
-f  and  — ,  signs  of  operation.     The  two  sets  of  signs  must,  as 
yet,  be  carefully  distinguished. 

30.  The  Absolute  Value  of  a  number  is  the  number  of  units 
•contained  in  it  without  regard  to  their  quality. 

E.g.,  the  absolute  value  of  +4  is  4,  of  ~5  is  5. 

31.  From  the  results  of  the  preceding  articles,  we  obtain  the 
following  general  relations : 

(i.)  Of  two  positive  numbers,  that  number  is  the  greater  which 
has  the  greater  absolute  value ;  and  that  number  is  the  less  which 
has  the  less  absolute  value. 

(ii.)  Of  two  negative  numbers,  that  number  is  the  greater  which 
has  the  less  absolute  value;  and  that  number  is  the  less  which  has 
the  greater  absolute  value. 

For  example,  ~3  >  ~5,  or  ~5  <  ~3,  since  ~5  is  five  units  less 
than  0,  and  ~3  is  only  three  units  less  than  0. 


27-34]         POSITIVE  AND  NEGATIVE   NUMBERS.  21 

32.  It  is  important  to  notice  that  a  negative  remainder  does 
not  mean  that  more  units  have  been  taken  from  the  minuend 
than  were  contained  in  it ;  such  a  remainder  indicates  that  the 
subtrahend  is  greater  than  the  minuend  by  as  many  units  as  are 
contained  in  the  remainder. 

Thus,  in  +15  —  +25  =  ~10,  the  remainder,  "10,  indicates  that 
the  subtrahend  is  10  units  greater  than  the  minuend. 

33.  It  is  evidently  necessary  thus  to  enlarge  the  meaning 
of  subtraction  in  such  an  expression  as  a  —  b.     For,  if  a  and  b 
are  to  have  any  values  whatever,  the  case  in  which  b  is  greater 
than  a,  that  is,  in  which  the  subtrahend  ft  greater  than  the  minu- 
end, must  be  included  in  the  operation  of  subtraction. 

34.  Negative  numbers  have  been  introduced  by  extending 
the  operation  of  subtraction.     But  it  is  necessary  to  treat  them 
as  numbers  apart  from  this  particular  operation. 

As  in  Arithmetic,  so  in  Algebra,  any  integer  is  an  aggregate 
of  like  units. 

Just  as  4  =  1  +  1  +  1  +  1, 

so       +4  =+1  ++1  ++1  ++1,  and  ~4  =~1  +~1  +~1  +~1. 

Just  as  |  =  i  + 1, 

so  +(|)  =+(J)  ++Q),  and  "(I)  =-(i)  +-(i). 

EXERCISES   VII. 

Simplify  the  following  expressions  : 

1.   +17- +4.               2.    +17 -+17.  3.  +17 -+27. 

4.   +25 -+18.             5.   +25-+25.  6.  +25  -+35. 

7.   +88 -+95.             8.   +56- +27.  9.  +101  - +105. 

What  value  of  x  will  make  the  first  member  of  each  of  the 
following  equations  the  same  as  the  second  ? 

10.    #-+5  =+7.         11.    x  -+5  =  0.  12.   a;— +5  =-2. 

13.   a -+11  =+9.       14.    a -+15  =-13.       15.   #-+12  =-10 


•2'2  ALGEBRA.  [Cn.  I 

How  many  units  is  each  of  the  following  numbers  greater  or 
less  than  0  ? 

16.    +5.          17.    -3.          18.    -11.          19.    +14.  20.    -20. 

Which  of  each  of  the  following  pairs  of  numbers  is  the 
greater,  and  by  how  many  units  ? 

21.   +7,  +4.          22.   ~7,  ~4.          23.   +7,  -4.          24.    ~7,  +4. 
25.   +19,  +9.         26.   -29,  +1.        27.   +32,  ~3.        28.    -14,  -3. 

Positive  and  Negative  Numbers  are  Opposite  Numbers. 

35.  In  Arithmetic  we  have :  the  remainder  added  to  the  sub- 
trahend is  equal  to  the  minuend.     This  principle,  like  all  princi- 
ples of  Arithmetic,  is  retained  in  Algebra.     We  therefore  have 
from  (iii.)  Art.  26 : 

+3  +3  +3  +3  +3  +3    subtr. 

+2  +J.  _0  2!  I?  I?    rem- 

+5  +4  +3  +2  +1  0     min. 

36.  The  equation  +3  -f  ~3  =  0  gives  us  the  following  impor- 
tant principle : 

TJie  sum  of  a  positive  number  and  a  negative  number  having 
the  same  absolute  value  is  equal  to  zero;  i.e.,  two  such  numbers 
cancel  each  other  ivhen  united  by  addition. 

E.g.,      +!+-!  =  0,  +3  +  ~3  =  0,  ~17J  +  +171  =  0. 
In  general,  +n  -f  ~n  =  0. 

For  this  reason,  positive  and  negative  numbers  in  their  rela- 
tion to  each  other  are  called  opposite  numbers.  When  their 
absolute  values  are  equal,  they  are  called  equal  and  opposite 
numbers. 

37.  Any  quantities  which  in  their  relation  to  each  other  are 
opposite,  may  be  represented  in  Algebra  by  positive  and  negative 
numbers  ;  as  credits  and  debits,  gain  and  loss. 


34-37]         POSITIVE   AND   NEGATIVE   NUMBERS.  23 

Ex.  1.  100  dollars  credit  and  100  dollars  debit  cancel  each 
other.  That  is,  100  dollars  credit  united  with  100  dollars  debit 
is  equal  to  neither  credit  nor  debit;  or, 

100  dollars  credit  +  100  dollars  debit  =  neither  credit  nor  debit. 

If  credits  be  taken  positively  and  debits  negatively,  then  100 
dollars  credit  may  be  represented  by  +100,  and  100  dollars  debit 
by  -100.  Their  united  effect,  as  stated  above,  may  then  be 
represented  algebraically  thus : 

+100 +-100  =  0. 

The  result,  0,  means  neither  credit  nor  debit. 
Similarly  for  opposite  temperatures. 

Ex.  2.  If  a  body  is  first  heated  10°  and  then  cooled  down  8°, 
its  final  temperature  is  2°  above  its  original  temperature ;  or, 

stated  algebraically, 

+10 +-8- +2. 

The  result,  +2,  means  a  rise  of  2°  in  temperature. 

EXERCISES  VIII. 

Express  algebraically  each  one  of  the  following  statements : 

1.  $  45  gain  and  $  45  loss  is  equivalent  to  neither  gain  nor 
loss. 

2.  $  95  gain  and  $  50  loss  is  equivalent  to  $  45  gain. 

3.  $  37  gain  and  $  57  loss  is  equivalent  to  $  20  loss. 

4.  If  a  man  travels  220  miles  due  west  and  then  220  miles 
due  east,  he  is  at  his  starting  place. 

5.  If  a  man  ascends  2250  feet  in  a  balloon  and  then  descends 
200  feet,  he  is  2050  feet  above  the  earth. 

6.  If  a  man  walks  90  feet  to  the  right  and  then  110  feet  to 
the  left,  he  is  20  feet  to  the  left  of  his  starting  point. 

7.  A  rise  of  20°  in  temperature,  fol-lowed  by  a  fall  of  27°,  is 
equivalent  to  a  fall  of  7°. 

8.  A  rise  of  15°  in  temperature,  followed  by  a  fall  of  12°,  is 
equivalent  to  a  rise  of  3°. 


CHAPTER   II. 

THE   FOUR    FUNDAMENTAL    OPERATIONS   WITH 
ALGEBRAIC    NUMBER. 

ADDITION   OF   ALGEBRAIC   NUMBERS. 

1.  TJie  Addition  of  two  numbers  is  the  process  of  uniting  them 
into  one  aggregate. 

The  numbers  to  be  added  are  called  Summands. 

Addition  of  Numbers  with  Like  Signs. 

2.  Ex.  1.    Add  +3  to  +4. 

The  three  positive  units,  +3,  when  added  to  the  four  positive 
units,  +4,  give  an  aggregate  of  four  plus  three,  or  seven,  positive 
units.  That  is, 

+4  +  +3  =+(4  +  3)  =+7. 

In  like  manner, 

Ex.2.    -4 +-3  =-(4 +  3)  =-7. 

These  examples  illustrate  the  following  method  of  adding 
two  numbers  with  like  signs : 

Add  arithmetically  their  absolute  values,  and  prefix  to  the  sum 
their  common  sign  of  quality. 

Addition  of  Numbers  with  Unlike  Signs. 

3.  Ex.  1.   Add  -2  to  +5. 

The  two  negative  units,  ~2,  when  added  to  the  five  positive 
units,  +5,  cancel  two  of  the  Jive  positive  units.  There  remain 
then  Jive  minus  two,  or  three,  positive  units.  That  is, 

+5+-2=+(o-2)=+3. 
24 


1-5]         SUBTRACTION   OF   ALGEBRAIC   NUMBERS.  25 

Ex.  2.    Add  +2  to  -5. 

The  two  positive  units,  +2,  when  added  to  the  five  negative 
units,  "5,  cancel  two  of  the  Jive  negative  units.  There  remain 
then  Jive  minus  two,  or  three,  negative  units.  That  is, 

-5  ++2  =-(5  -2)  =-3. 

Observe  that  in  both  examples  the  sum  is  of  the  same  quality 
as  the  number  which  has  the  greater  absolute  value.  Also,  that 
the  absolute  value  of  the  sum  is  obtained  by  subtracting  the  less 
absolute  value,  2,  from  the  greater,  5. 

•These  examples  illustrate  the  following  method  of  adding 
two  numbers  with  unlike  signs  : 

Subtract  arithmetically  the  less  absolute  value  from  the  greater. 
To  that  remainder  prefix  the  sign  of  quality  of  the  number  which 
has  the  greater  absolute  value. 

The  examples  given  in  Ch.  I,  Art.  37,  are  concrete  illustra- 
tions of  the  preceding  principles. 

4.  Observe  that  a  positive  number  increases  a  number  to 
which  it  is  added,  while  a  negative  number  decreases  it. 

EXERCISES  I. 
Add: 

1.  2.  3.  4.  5.  6. 

+2  -4  +9  -8  +13  -21 

+6  -5  +3  -7  +19  -15 

7.  8.  9.  10.  11.  12. 

+8  -8  -7  +13  -21  +37 

-3  +3  +4  -17  +32  -22 


SUBTRACTION  OF   ALGEBRAIC  NUMBERS. 

5.   Subtraction  is  the  inverse  of  addition.     In  addition  two 
numbers  are  given,  and  it  is  required  to  find  their  sum,  as 

in  +9  ++2  =+11. 


26  ALGEBRA.  [Cn.  II 

In  subtraction  the  sum  of  two  numbers  and  one  of  them  are 
given,  and  it  is  required  to  find  the  other  number,  as  in 

+11  -+2  =  (+9  ++2)  -+2  =+9. 

That  is,  if  from  the  sum  of  two  numbers  either  of  the  numbers 
be  subtracted,  the  remainder  is  the  other  number. 

In  general,  (a  +  6)  —  a  —  b. 

6.  Ex.  l.    A  man's  net  profits  last  year  were  1200  dollars. 
This  year  his  income  is  150  dollars  less,  and  his  expenditures 
are  the  same.     What  are  his  net  profits  this  year  ? 

To  take  aivay  150  dollars  income  is  equivalent  to  adding  150 
dollars  expenditures. 

If  net  profits  and  income  be  taken  positively,  and  expendi- 
tures negatively,  the  last  statement,  expressed  algebraically,  is 

+1200  -+150  =+1200  +-150. 

Ex.  2.  A  man's  net  profits  last  year  were  1200  dollars.  This 
year  his  income  is  the  same  and  his  expenditures  are  150  dol- 
lars less.  What  are  his  net  profits  this  year  ? 

To  take  away  150  dollars  expenditures  is  equivalent  to  adding 
150  dollars  profits. 

The  algebraic  statement  of  this  relation  is 

+1200  - -150  =+1200  ++150. 
These  examples  illustrate  the  following  principle : 

To  subtract  one  number  from  another  number,  reverse  the  sign 
of  quality  of  the  subtrahend,  and  add. 

E.g.9   +2  -+3  =+2  +-3,  =-1.          -2  -+3  =~2  +~3  =~5. 
+2  -~3  =+2  ++3,  =+5.          -2  -~3  =~2  ++3  =  +1. 

7.  It  is  important  to  notice  that  the  preceding  examples  do 
not  prove  this  principle.     The  following  examples  illustrate  a 
method  of  proof  which  may  be  used. 


5-9]     MULTIPLICATION  OF   ALGEBRAIC   NUMBERS.        27 

Ex.  1.    Subtract  +5  from  +7. 

In  +7  —  +o,  the  minuend,  +7,  is  to  be  expressed  as  the  sum  of 
two  numbers,  one  of  which  is  +5.  Since  ~5  H-+5  =  0,  we  may 

write 

+7  =+7  +-5  ++5  =  (+7  +-5)  ++5. 

That  is,  +7  may  be  regarded  as  the  sum  of  two  numbers,  one 
of  which  is  +7  +  ~5,  and  the  other  is  +5.  Therefore,  by  defini- 
tion of  subtraction, 

+7  -+5  =  [(+7  +-5)  ++5]  -+5 

=  +7  +-5  =+2, 

That  is,  to  subtract  +5  is  equivalent  to  adding  ~5. 
Ex.  2.    Subtract  ~5  from  +7. 

We  have         +7  -~5  =  [(+7  ++5)  +~5]  -~5 
=  +7  ++5  =+12, 

That  is,  to  subtract  ~5  is  equivalent  to  adding  +5. 

8.  We  thus  see  that  every  operation  of  subtraction  is  equiva- 
lent to  an  operation  of  addition.  On  this  account  it  is  conven- 
ient to  speak  of  a  chain  of  additions  and  subtractions  as  an 
Algebraic  Sum. 

EXERCISES  II. 
Subtract  : 

1.  2.  3.  4.  5.  6. 

+9  +2  +8  +3  -9  -4 

+2  +9  +3  +8  -4  -9 

7.  8.  9.  10.  11.  12. 

-8  -7  +5  -6  +6  -(> 


MULTIPLICATION  OF  ALGEBRAIC   NUMBERS. 

9.   In  multiplication,  the  multiplicand  and  multiplier  are 
called  Factors  of  the  product. 


28  ALGEBRA.  [Cn.  n 

10.  In   ordinary  Arithmetic,  multiplication  by  an  integer 
is  denned  as  an  abbreviated  addition.     Thus, 

4  x3  =  4  +  4  +  4; 
that  is,  the  number  4  is  taken  three  times  as  a  summand. 

But  3  =  1  +  1  +  1. 

We  thus  see  that  the  product  4  x  3  is  obtained  from  4  just 
as  3  is  obtained  from  the  positive  unit,  1. 

We  are  thus  naturally  led  to  the  following  definition  of 
multiplication  : 

The  product  is  obtained  from  the  multiplicand  just  as  the  mul- 
tiplier is  obtained  from  the  positive  unit. 

11.  The  above  definition  is  an  extension  of  the  meaning  of 
arithmetical  multiplication  when  the  multiplier  is  an  integer, 
and  gives  an  intelligible  meaning  to  arithmetical  multiplication 
when  the  multiplier  is  a  fraction. 

Thus,  |-  is  obtained  from  the  unit,  1,  by  taking  one-third  of 
the  latter  twice  as  a  summand  ;  or 


In  like  manner,  to  multiply  5  by  -f,  we  take  one-third  of  5 
twice  as  a  summand  ;  or 


12.   There  are  two  cases  to  be  considered  in  the  multiplica- 
tion of  algebraic  numbers. 

(i.)  The  Multiplier  Positive.  —  Ex.  1.    Multiply  +4  by  +3. 
By  the  definition  of  multiplication,  the  product, 

+4  x  +3, 

is  obtained  from  +4  just  as  +3  is  obtained  from  the  positive 
unit.     But 

+3=+l++l++l. 

Consequently  the  required  product  is  obtained  by  taking  +4 
three  times  as  a  summand,  or 

+4  x  +3  =+4  ++4  ++4  =+(4  +  4  +  4)  =+(4  x  3)  =+12. 


10-13]     MULTIPLICATION   OF   ALGEBRAIC   NUMBERS.    29 

Ex.  2.    Multiply  -4  by  +3. 
By  the  definition  of  multiplication,  we  have 
-4  x  +3  =~4  +-4  +~4  =-(4  +  4  +  4)  =~(4  x  3)  =-12. 

(ii.)  The  Multiplier  Negative.  —  Ex.  3.    Multiply  +4  by  ~3. 
By  the  definition  of  multiplication,  the  product, 

+4x-3, 
is  obtained  from  +4  just  as  ~3  is  obtained  from  the  positive 

unit.     But 

-3  =~1  +-1  +-1  =  -+1  -+1  -+1 ; 

that  is,  ~3  is  obtained  by  subtracting  the  positive  unit,  +1,  three 
times  in  succession  from  0.  Consequently,  the  required  product 
is  obtained  by  subtracting  the  multiplicand,  +4,  three  times  in 
succession  from  0 ;  or, 

+4  x-3  =  -+4-+4-+4  =  +-4+-4+-4=-(4  x  3). 

Ex.  3.   Multiply  ~4  by  ~3. 

By  the  definition  of  multiplication,  we  have 

-4  x  -3  =  --4  --4  --4  =  ++4  ++4  ++4  =+(4  x  3). 

13.  These  examples  illustrate  the  following  Rule  of  Signs 
for  Multiplication: 

The  product  of  two  numbers  having  like  signs  is  positive ;  and 
the  product  of  two  numbers  having  unlike  signs  is  negative.  Or, 
stated  symbolically, 

+a  x+b  =+(ab\  ~a  x+6  =-(a6), 

-a  x  ~b  =+(ab\  +a  x  ~b  =-(a6). 

EXERCISES  III. 


Multiply  : 

1. 

2. 

3. 

4. 

5. 

6. 

+3 

-3 

+3 

-3 

+8 

-7 

+4 

+4 

Ii 

^4 

£5 

-6 

7. 

8. 

9. 

10. 

11. 

12. 

-9 

+8 

-12 

-15 

+2Q 

+16 

+2 

-6 

-5 

+4 

+7 

~5 

30  ALGEBRA.  [Cn.  II 

DIVISION   OF   ALGEBRAIC    NUMBERS. 

14.  Division  is  the  inverse  of  multiplication.     In  multiplica- 
tion two  factors   are  given,  and   it  is   required  to  find  their 
product.     In  division  the  product  of  two  factors  and  one  of 
them  are  given,  and  it  is  required  to  find  the  other  factor. 

E.g.,  Since  -28=-4x+7, 

therefore,          ~28  -h+7  =  ~4,  and  "28  -j-~4  =  +7. 

15.  From  the  definition  of  division  we  infer  the  following 
principle : 

If  the  product  of  two  numbers  be  divided  by  either  of  them,  the 
quotient  is  the  other  number. 

16.  Since  +ax+b=+(ab),  therefore  +(ab)  -=-+a  =+b; 
since  ~a  x+6=~(a6),  therefore  ~(ab)  ^-~a  =+6; 
since  ~a  x~b=+(ab),  therefore  +(a6)  -5-~a  =~b; 
since  +a  x~b=~(ab),  therefore  ~(ab)  -r-+a  =~6. 

From  these  equations  we  derive  the  following  Rule  of  Signs 
for  Division : 

Like  signs  of  dividend  and  divisor  give  a  positive  quotient; 
unlike  signs  of  dividend  and  divisor  give  a  negative  quotient. 

E.g.,  +8 -=-+2  =+4;  -8-=--2=+4; 

+8-s--2=-4;  -8 -s- +2  =-4. 


EXERCISES  IV. 

Divide : 
1.  2.  3.  4. 

+4)^8  +4)^8  -4)+8  -4)2 

6.  7.  8.  9.  10. 

~7)+28  +6)-30  +8)+24 


14-19]        SIGNS   OF   QUALITY  AND   OPERATION.  31 

ONE   SET   OF   SIGNS   FOR   QUALITY  AND   OPERATION. 

17.  Most  text-books  of  Algebra  use  the  one  set  of  signs, 
+  and  — ,  to  denote  both  quality  and  operation.     We  shall  in 
subsequent  work  follow  this  custom.     For  the  sake  of  brevity, 
the  sign  +  is  usually  omitted  when  it  denotes  quality ;    the 
sign  —  is  never  omitted. 

Thus,  instead  of  +2,  we  shall  write  +  2,  or  2 ; 
instead  of  ~2,  we  shall  write  —  2. 

18.  We  have  used  the  double  set  of  signs  hitherto  in  order 
to  emphasize  the  difference  between  quality  and  operation.     It 
should  be  kept  clearly  in  mind  that  the  same  distinction  still 
exists. 

We  now  have 

+3  _|_+2  =  +  3  +  (+  2)  =  3  +  2,  omitting  the  signs  of  quality,  + ; 

+3  _l_-2:= -}-3 +(— 2),  wherein  +  denotes  operation,  and  —  de- 
notes quality. 

+3  _+2  =  +  3  —  (+  2)=  3  —  2,  omitting  the  signs  of  quality,  +  ; 

+3  _~2  =  +  3  —  (—  2),  wherein  the  first  sign,  — ,  denotes  opera- 
tion, the  second  sign,  — ,  denotes  quality. 

19.  In  the  chain  of  operations 

(+2)  +  (-5)-(+2)-(-ll) 

the  signs  within  the  parentheses  denote  quality,  those  without 
denote  operation.     That  expression  reduces  to 

(+  2)  -(+5)  -(+2)  +  (+H), 
or  2-5-2  +  11, 

dropping  the  sign  of  quality,  +. 

In  the  latter  expression  all  the  signs  denote  operation,  and 
the  numbers  are  all  positive. 


32  ALGEBRA.  [Cn.  II 

20.  The  following  examples  illustrate  the  double  use  of  the 
signs  +  and  —  . 

Ex.1.  +4  ++3  =  +  4  +  (+3)  =  4  +  3  =  7. 

Ex.2.  -5++2  =  -5  +  (+2)  =  -5  +  2  =  -3. 

Ex.  3.  +7  --5  =  +  7  -  (-  5)  =  7  -  (-  5)  =  7  +  5  =  12. 

Ex.4.  ~4x+3  =  -4x  (+3)=-4x3  =  -12. 

Ex.5.  -4x-3  =  -4x(-3)  =  12. 

Continued  Products. 

21.  The  results  of  Article  13  may  be  applied  to  determine 
the  value  of  a  chain  of  indicated  multiplications,  i.e.,  of  a 
continued  product. 

E.g.         (+  a)  (+  6)  (+  C)  =  (+  ab)  (+  e)  =  +  abc, 

(+  a)  (+  6)  (-  G)  =  (  +  ab)  (-c)  =  -  abc, 
(+  a)  (-  6)  (-  c)  =  (-  a6)  (-  c)  -  +  abc, 


These  equations  illustrate  a  more  general  rule  of  signs  : 

A  continued  product  which  contains  no  negative  number,  or  an 
even  number  of  negative  numbers,  is  positive;  one  that  contains 
an  odd  number  of  negative  numbers  is  negative. 

In  practice  the  sign  of  a  required  product  may  first  be  deter- 
mined by  inspection,  and  that  sign  prefixed  to  the  product  of 
the  absolute  values  of  the  numbers  in  the  continued  product. 

E.g.,  the  sign  of  the  product 

2x(-3)x(-7)x(+4)x(-5) 

is  negative,  since  it  contains  three  negative  numbers  ;  the  product 
of  the  absolute  values  is  840.     Consequently, 

2x  (-3)  x  (-7)x  (+4)  x  (-5)  =  -840. 


20-21]        SIGNS   OF   QUALITY   AND   OPERATION.  33 

EXERCISES   V. 

In  the  expressions  in  Exx.  1-4,  which  signs  denote  quality 
and  which  operation  ? 

1.    +5 -}- (-3) -(+8).  2.    .-7 +  (+5) -(-9). 

3.    _3  +  (-5)x(+4).  4.    (+12)H-(-4)x(-3). 

5-8.    Find  the  value  of  the  expressions  in  Exx.  1-4. 

Find  the  values  of  the  expressions  in  Exx.  9-20,  first  chang- 
ing them  into  equivalent  expressions  in  which  there  is  only 
one  set  of  signs  -f  and  — : 

9.   +8 +  +2.      10.   +7-+3.      11.   +3- +7.       12.   ~5  +  ~7. 
13.    -8  -  +3.      14.    -9  -  -5.       15.    +4  x  +5.       16.    +5  x  ~2. 
17.   ~5  x  ~2.      18.   +12  -*-  +3.     19.   +12  -f-  -3.     20.   -12  -r-  -3. 
Simplify  the  following  expressions  : 

21.    10-4.  22.    4-10.  23.  -8-7. 

24.    9-2.  25.    2-9.  26.  -10  +  10. 

27.  8x5.  28.  -  8  x  5.  29.   8  x  (-  5). 

30.  (_8)x(— 5).    31.  20 -=-4.  32.    —20-*- 4. 

33.  20  -s- (—4).        34.  (-20)-f-(-4).    35.    -45-T-9. 

36.  3  x  5  4-  4  x  2.  37.   3  x  (5  +  4  x  2). 

38.  8  x  6  -  10  -3-  5.  39.    (8  x  6  -  10)  -*-  5. 

40.  12 -s- 4  —  10 -5- 2.  41.   12 -j- (4  -  10 -5- 2). 

When  a  =  16,  b  =  -  8,  c  =  -  2,  d  =  -  4,  find  the  values  of : 
42.    a  +  b  -f-  c.  43.    a  +  b  —  c.  44.    a  —  6  -f-  c. 

45.    a  —  b  —  c.  46.    a  —  (b  —  c).  47.    c  —  (6  —  a). 

48.    a&c.  49.    ab  -s-  c.  50.   a  -*-  (6c). 

51.    a  -=-  b  x  c.  52.    abed.  53.    (aft)  -f-  (cd). 

54.    abc  -r-  d.  55.    ab  +  cd.  56.    a  -s-  b  —  d  -=-  c. 

57.  A's  assets  are  $  2600  and  B's  are  $  2200.  How  much 
do  A's  assets  exceed  B's,  taking  assets  positively  ? 


34  ALGEBRA.  [Cn.  II 

58.  A  owes  $200,  and  B's  assets  are  $1800.     How  much 
do  A's  assets  exceed  B's,  taking  assets  positively  ? 

59.  The  temperature  in  a  room  is  72°  above  zero,  and  out 
of  doors  it  is  8°  above  zero.     How  much  higher  is  the  tem- 
perature in  the  room  than  out  of  doors,  taking  degrees  above 
zero  positively? 

60.  The  temperature  in  a  room  is  70°  above  zero,  and  out 
of  doors  it  is  4°  below  zero.     How  much  higher  is  the  tem- 
perature in  the  room  than  out  of  doors,  taking  degrees  above 
zero  positively  ? 

PARENTHESES. 

22.  The  Terms  of  an  algebraic  sum  are  the  additive  and  sub- 
tractive  parts  of  tne  sum. 

E.g.,  the  terms  of  2  —  5  —  2  +  11  are  +  2,  —  5,  —  2,  +  11 

The  Sign  of  a  Term  is  its  sign  +  or  — . 

A  Positive  Term  is  one  whose  sign  is  +  ;  as  +2. 

A  Negative  Term  is  one  whose  sign  is  —  ;  as  —  5. 

Removal  of  Parentheses. 

23.  We  have       9  +  (5  +  6)  =  9  +  5  +  6, 

since  to  add  the  sum  5  +  6  is  equivalent  to  adding  successively 
the  single  numbers  of  that  sum. 

Again,  9  +  (5  -  6)  =  9  +  [5  +  (-  6)], 

since  to  add  —  6  is  equivalent  to  subtracting  6. 
Therefore,  removing  brackets, 

9  +  (5  -  6)  =  9  +  5  +  (-  6),  =9  +  5-6. 
The  above  example  illustrates  the  following  principle : 

(i.)  When  the  sign  of  addition,  -f,  precedes  parentheses,  they 
may  be  removed,  and  the  signs,  +  and  —>  within  them  be  left 
unchanged;  that  is, 

yy+(+fl  +  6)  =  /Y  +  a  +  6, 
N+(+a-b)=N  +  a-b,  etc. 


21-25]  PARENTHESES.  35 

It  is  important  to  notice  that  if  the  first  term  within  the 
parentheses  has  no  sign,  the  sign  +  is  understood. 

24.  We  also  have 

9-(5  +  6)  =  9-5-6, 

since  to  subtract  the  sum  5  +  6  is  equivalent  to  subtracting 
successively  the  single  numbers  of  that  sum. 

Again,  9  -  (5  -  6)  =  9  -  [5  +  (-  6)], 

since  to  add  —  6  is  equivalent  to  subtracting  6. 
Therefore,  removing  brackets, 

9  _  (5  _  6)  =  9  -  5  -  (-  6),  =9-5  +  6. 
This  example  illustrates  the  following  pri^-iple: 

(ii.)  When  the  sign  of  subtraction,  —  ,  precedes  parentheses, 
they  may  be  removed,  if  the  signs  within  them  be  reversed  from  + 
to  _?  and  from  —  to  +  ;  that  is, 

yy-(+  fl  +  6)  =  yv-a-6, 

N-(+a-b)  =  N-a  +  b,  etc. 

Observe  that  the  sign  before  the  parentheses  affects  each 
term  within  them. 

Insertion  of  Parentheses. 

25.  The  insertion  of  parentheses   is   the  converse   of  the 
process  of  removing  them. 

(i.)  An  expression  may  be  inclosed  within  parentheses  preceded 
by  the  sign  +  ,  if  the  signs  of  the  terms  inclosed  remain  unchanged. 

E.g.,  7-5  +  3  -4  =  7  +  (-5  +  3-  4), 

=  7-5 


(ii.)  An  expression  may  be  inclosed  within  parentheses  pre- 
ceded by  the  sign  —,  if  the  signs  of  the  terms  inclosed  be  reversed, 
from  +  to  —  and  from  —  to  +. 

E.g.,  7-5  +  3-4  =  7-  (5  -3  +  4), 

=  7-5-(-3 


36  ALGEBRA.  [Cn.  II 

EXERCISES  VI. 

Find  the  value  of  each  of  the  following  expressions,  first 
removing  parentheses  : 

1.   9  +  (4+3).  2.   9  +  (4-3).  3.   10-(3+4). 

4.   io_(3-4).  5.   12+  (6+8).  6.   12-(6+8). 

7.   i2_(6-8).  8.   12  +  (-6+8).         9.   12-(-6+8). 

10.   15  +  (9-6+2).      11.   15-(9-6+2).     12.   20-(7-9-l). 
13.    is      _45-8.  14.   l8--45-8. 


Insert  parentheses  in  10  —  7  +  4  —  6  and  7  +  8  —  9  —  4, 

15.  To  inclose  the  last  two  terms,  preceded  by  the  sign  +  ; 
preceded  by  the  sign  —  . 

16.  To  inclose  the  last  three  terms,  preceded  by  the  sign  +  ; 
preceded  by  the  sign  —  . 

The  Associative  Law. 

26.  The  principle  for  inserting  parentheses  enables  us  to 
group  successive  terms  in  algebraic  addition. 

Kg.,     8  +  (4  +  l)  =  (8  +  4)  +  l,  or  8  +  5  =  12  +  1. 
In  general,          a  +  (b  +  c)  =  (a  +  6)  +  c. 

That  is,  the  algebraic  sum  of  three  or  more  numbers  is  the 
same  in  whatever  way  successive  numbers  are  grouped  or  asso- 
ciated in  the  process  of  adding.  . 

This  principle  is  called  the  Associative  Law  for  addition  and 
subtraction. 

27.  In  finding  the  value  of  a  continued  product  in  Art.  21  9 
the  indicated  operations  were  performed  successively  from  left 
to  right. 

E.g.,  4  x  3  x  (-  2)  =  12  x  (-  2)  =  -  24. 

But  the  result  will  be  the  same  if  3  be  first  multiplied  by 
—  2,  and  then  4  be  multiplied  by  this  product. 


25-29]  PARENTHESES.  37 

E.g.,  4  x  [3  x  (-2)]  =  4  x  (-  6)  =  -24. 

In  general,  (ab)c  =  a  (be). 

That  is,  the  product  of  three  or  more  numbers  is  the  same  in 
whatever  way  two  or  more  successive  numbers  are  grouped  or 
associated  in  the  process  of  multiplying. 

This  principle  is  called  the  Associative  Law  for  multiplication. 

The  Commutative  Law. 

28.  In  an  indicated  addition,  the  number  on  the  right  of 
the  sign  +  is  to  be  added  to  the  number  on  the  left. 

E.g.,  in  5  +  3,  =8, 

3  is  added  to  5 ;  while  in   3  +  5,  =8, 
5  is  added  to  3.     But  the  results  are  the  same.     That  is, 

5+3=3+5 

In  like  manner,  8  —  5  =  —  5  +  8. 

In  general,  a  +  b  =  b  +  a ; 

a  +  6  —  c  =  a  —  c  +  b  =  etc. 

That  is,  the  algebraic  sum  of  two  or  more  numbers  is  the  same 
in  whatever  order  they  may  be  added. 

This  principle  is  called  the  Commutative  Law  for  addition 
and  subtraction. 

29.  In  an  indicated  multiplication,  the  number  which  follows 
the  symbol  of  multiplication  is  the  multiplier. 

E.g.,  in  4x3  =  4+4  +  4  =  12, 

the  multiplier  is  3 ;  while  in 

3x4  =  3  +  3  +  3  +  3  =  12, 
the  multiplier  is  4.     But  the  results  are  the  same.     That  is, 

4x3  =  3x4. 
In  general,  a  x  b  =  b  x  a  ; 

a  x  b  x  c  =  a  x  c  x  b  =  etc. 


38  ALGEBRA.  [Cn.  II 

That  is,  the  product  of  two  or  more  numbers  is  the  same  in 
whatever  order  they  may  be  multiplied. 

This  principle  is  called  the  Commutative  Law  for  multiplica- 
tion. 

30.  By  the  preceding  articles  we  have : 

8-3  +  2-5  =  8  +  2-3-5=   10-   8  =  2  (i.) 

25  x   27   x  4  =  25  x  4  x  27  =  100  x  27  =  2700,       (ii.) 
75  x  29  -r-  25  =  75  -r-  25  x  29  =      3  x  29  =  87.          (iii.) 

In  changing  the  order  of  the  operations,  it  is  important  to 
carry  the  symbol  of  operation  with  the  number. 

31.  Thus,   by  the   methods   of   the    preceding   article,    we 
secure  the  following  advantages: 

In  a  succession  of  additions  and  subtractions,  add  the  positive 
terms  separately,  then  the  negative  terms,  and  unite  the  results, 
as  in  (i.). 

In  a  succession  of  multiplications  and  divisions,  we  may,  by 
changing  the  order  of  the  operations,  frequently  simplify  the 
work,  as  in  (ii.)  and  (iii.). 

EXERCISES   VII. 

Find  the  value  of  each  of  the  following  expressions : 
1.   8-3  +  2-5  +  9.  2.    -6  +  4-14  +  12-7. 

3.    i9_7  +  3_5_io.  4.    16-7  +  4-9  +  3. 

5.    17  +  2-3  +  9  -18.  6.    15-19  +  6-7  +  5. 

Find,  in  the  most  convenient  way,  the  value  of  each  of  the 
following  expressions : 

7.   89-115  +  11.  8.  45f-85  +  54f. 

9.  996  +  1008  +  4-8.  10.  98  +  96  +  92  +  2  +  4  +  8. 

11.  25  x  32  x  (-  4).  12.  12|  x  (-  29)  x  8. 

13.  -  39  x  16|  x  6.  14.  45  x  28  -h  9. 

15.  -  121  -j-  20  x  8.  16.  10  --  42  x  21. 


29-37]  POSITIVE   INTEGRAL  POWERS.  39 

POSITIVE   INTEGRAL  POWERS. 

32.  The  Sign  of  Continuation,  •  ••,  is  read,  and  so  on,  or  and  so 
on  to;  as  1,  2,  3,  •••,  read,  one,  two,  three,  and  so  on;  or  1,  2,  3, 
•  ••,  10,  read,  one,  two,  three,  and  so  on  to  10. 

33.  A  continued  product  of  equal  factors  is  called  a  Power 
of  that  factor. 

Thus,  2  x  2  is  called  the  second  power  of  2,  or  2  raised  to  the 
second  power ;  aaa  is  called  the  third  power  of  a,  or  a  raised  to 
the  third  power. 

In  general  aaa  •••  to  n  factors  is  called  the  nth  power  of  a, 
or  a  raised  to  the  nth  power. 

The  second  power  of  a  is  often  called  the  square  of  a,  or  a 
squared;  and  the  third  power  of  a  the  cube  of  a,  or  a  cubed. 

34.  The  notation  for  powers  is  abbreviated  as  follows : 

a2  is  written  instead  of  aa ;  a3  instead  of  aaa ; 
an  instead  of  aaa  -  •  •  to  w  factors. 

35.  The  Base  of  a  power  is  the  number  which  is  repeated  as 
a  factor. 

E.g.,  a  is  the  base  of  a2,  a3,  •••,  a". 

36.  The  Exponent  of  a  power  is  the  number  which  indicates 
how  many  times  the  base  is  used  as  a  factor,  and  is  written  to 
the  right  and  a  little  above  the  base. 

E.g.,  the  exponent  of  a2  is  2,  of  a3  is  3,  of  an  is  n. 
The  exponent  1  is  usually  omitted.     Thus,  a1  =  a. 

37.  The  base  of  a  power  must  be  inclosed  within  parentheses 
to  prevent  ambiguity : 

(i.)    When  the  base  is  a  negative  number.     Thus, 
(_5)2=(-5)(-5)  =  25;  while  -  52  =  -  (5  x  5)  =  -  25. 


40  ALGEBRA.  [Cn.  II 

(ii.)  When  the  base  is  a  product  or  a  quotient.     Thus, 

(2  x  5)3=  (2  x  5)(2  x  5)(2  x  5)  =  1000; 
while  2  x  53  =  2  (5  x  5  x  5)  =  250. 

:.*„,.  (f)'=|xf4  <*>.f-4-»-g. 

(iii.)    When  the  base  is  a  sum.     Thus, 


while  2  +  32  =  2  +  3x3  =  2  +  9  =  11. 

(iv.)  When  the  base  is  itself  a  power.     Thus, 

(23)2  =  23  x  23  =  (2  x  2  x  2)  (2  x  2  x  2)  =  64. 
while  2s2  =  23X3  =29  =  2x2x2x2x2x2x2x2x2  =  512. 

EXERCISES   VIII. 

Express  each  of  the  following  powers  in  the  abbreviated 
notation : 

1.   a  x  a.        2.   4  x  4.        3.   2  x  2  x  2.        4.    (-  a)  (-  a). 
5.    -ax  a.         6.    (-3)  (-3)  (-3)  (-3).         7.    -nnnnn. 

8.  2  x  2  x  2 ...  to  8  factors. 

9.  (—  a)  (—  a) (—  a)  .«.  to  9  factors. 

10.    (a +  6)  (a +  6)  (a +  6).          11.    (x-yy)(x-yy)(x-yy). 
12.    (a  +  6)  (a  +  6)  (a  +  6)  ••  •  to  12  factors. 

Express  each  of  the  following  powers  as  a  continued  product : 
13.    36.  14.    63.  15.     -43.  16.    (— 4)8. 

17.    xf.  18.    (xy)s.  19.   (-a)4.  20.    -a4. 

Write : 

21.    Four  times  #.  22.   x  to  the  fourth  power. 

23.  The  sum  of  the  cubes  of  a  and  b. 

24.  The  cube  of  the  sum  of  a  and  b. 

25.    The  length  of  a  side  of  a  square  floor  is  a  feet.     How 
many  square  feet  in  the  floor  ? 


37-38]  POSITIVE  INTEGRAL  POWERS.  41 

26.  A  field  is  3  a  rods  long  and  2  a  rods  wide.     How  many 
square  rods  in  its  area  ? 

27.  A  box  is  4  #  feet  long,  3  x  feet  wide,  and  2  x  feet  high. 
How  many  cubic  feet  does  it  contain  ? 

Properties  of  Positive  Integral  Powers. 

38.   (i.)  All  (even  and  odd)  powers  of  positive  bases  are  positive. 
E.g.,  23  =  2x2x2  =  8.  34  =  3  x  3  x  3  x  3  =  81. 

(ii.)  Even  poivers  of  negative  bases  are  positive;  odd  powers 
of  negative  bases  are  negative. 

E.g.,  (-  2)4  m  (-  2)  (-  2)  (-  2)  (-  2)  =  16 ; 

(_  5)3  =(-5)  (-5)  (-5)  =  -125. 
In  general,  (  +  a)m  =  +  am ; 

(  —  ay2"  =  a1" ;    (  —  a)2w+1  =  —  a2w+1. 

EXERCISES   IX. 

Find  the  value  of  each  of  the  following  powers : 
1.   2'5.  2.   52.          3.    (-2)6.      4.    -26.  5.    (-3)5. 

6.    (-2)8.      7.    -33.     8.    (-3)3.       9.    (-a)6.       10.    (-a)9. 

Express  as  powers  of  2  : 

11.   8.      "  12.   32.         13.   128.        14.   1024.         15.   4096. 

Express  as  powers  of  —  3 : 

16.   9.       17.    -27.       18.    -243.       19.    729.       20.    -2187. 

Find  the  value  of  each  of  the  following  expressions  : 

21.    22  +  32.          22.    (2  +  3)2.      23.    33-23.  24.    (3-2)3. 

25.    (4x3)2.       26.   6x42.        27.   2(-3)3.       28.    [2(-3)]3. 

When  a  =  5,  b  =  —  4,  c  =  2,  find  the  value  of  each  of  the 
following  expressions : 

29.    ac.          30.    ba.          31.    (ab)c.          32.    bca.          33.    (abc)e. 
34.      a_£_c2.  35.    a2_62_c2_  36       a.2 


CHAPTER   III. 

THE    FUNDAMENTAL     OPERATIONS     WITH    INTEGRAL 
ALGEBRAIC   EXPRESSIONS. 

DEFINITIONS. 

1.  An  Integral  Algebraic  Expression  is  an  expression  in  which 
the  literal  numbers  are  connected  only  by  one  or  more  of  the 
symbols  of  operation,  +,  —  ,  x,  but  not  by  the  symbol  -;-. 

E.g.,  1  -f  x  +  x-j  o  a2b  +  f  cd2,  etc. 

2.  The  word  integral  refers  only  to  the  literal  parts  of  the 
expression. 

E.g.,  a  -f-  b  is  algebraically  integral  ;  but  when  a  =  ^,  6  =  f  , 
we  have 


3.  Coefficients.  —  In  a  product,  any  factor,  or  product  of  fac- 
tors, is  called  the  Coefficient  of  the  product  of  the*  remaining 
factors. 

E.g.,  in  3  abc,  3  is  the  coefficient  of  abc,  3  b  of  ac,  etc. 
A  Numerical  Coefficient  is  a  coefficient  expressed  in  figures. 
E.g.,  in  —  Sab,  —  3  is  the  numerical  coefficient  of  ab. 
A  Literal  Coefficient  is  a  coefficient  expressed  in  letters,  or  in 
letters  and  figures. 

E.g.,  in  3  ab,  a  is  the  literal  coefficient  of  3  b,  and  3  a  of  6, 
The  coefficients  +  1  and  —  1  are  usually  omitted. 

4.  A  coefficient  must  not  be  confused  with  an  exponent. 

E.g.,  4a  =  aH-a  +  a-fa;  while  a4  =  a  x  a  x  a  x  a. 

42 


1_7]  DEFINITIONS.  43 

5.  The  sign  +,  or  the  sign  —  ,  preceding  a  product,  is  to  be 
regarded  as  the  sign  of  its  numerical  coefficient. 

Thus  +  3  a  means  the  product  of  positive  3  by  a  ;  —  5  x 
means  the  product  of  negative  5  by  x.  In  particular,  -f-  a  means 
the  product  of  positive  1  by  a,  and  —  a  means  the  product  of 
negative  1  by  a,  unless  the  contrary  is  stated. 

EXERCISES  I. 

What  is  the  coefficient  of  x  in 
l.   2x?  2.    -3x?  3.   5  ax?  4.    -7bx? 

5.  If  the  sum,  a  -f  a  -f  a  -f-  a,  be  represented  as  a  product, 
what  is  the  coefficient  of  a  ? 

6.  If  the  algebraic  sum,  —  b  —  b  —  b  —  b  —  b,  be  represented 
as  a  product,  what  is  the  coefficient  of  —  b  ?     Of  b? 

7.  If  the  sum  ax  -f  ax  +  ax  -f-  •••  to  10  terms  be  represented 
as  a  product,  what  is  the  coefficient  of  ax  ?     Of  x? 

6.  Like  or  Similar  Terms  are  terms  which  do  not  differ,  or 
which  differ  only  in  their  numerical  coefficients. 

E.g.,  in  the  expression  +  3  a  +  6  a&  —  5a  +  7  ab,  -h  3  a  and 
—  5  a  are  like  terms  ;  so  are  +  6  ab  and  -f-  7  ab. 

Unlike  or  Dissimilar  Terms  are  terms  which  are  not  like. 
.g.9  -f  3  a  and  —  7  ab  in  the  above  expression. 


7.   A  Monomial  is  an  expression  of  one  term  ;  as  a,  —  7  be. 
A  Binomial  is  an  expression  of  two  terms  ;  as  —  2  a2  +  3  be. 
A  Trinomial  is  an  expression  of  three  terms. 
E.g.,  a  +  b-c,  -  3  a2  +  7  b*  -  5  c4. 

A  Multinomial  *  is  an  expression  of  two  or  more  terms,  in- 
cluding, therefore,  binomials  and  trinomials  as  particular  cases. 

E.g.,  a  +  b2,  a2  +  b-  c3,  ab  +  bc-cd-  ef. 

*  The  word  Polynomial  is  frequently  used  instead  of  Multinomial. 


44  ALGEBRA.  [On.  Ill 

ADDITION  AND   SUBTRACTION. 
Addition  of  Like  Terms. 

8.   Like  Terms  can  be  united  by  addition  into  a  single  like 
term. 

Just  as  2  =  1  +  1,  so  2  xy  =  xy  +  xy  ; 

just  as  3  =  1  +  1  +  1,  so  3  xy  =  xy  +  ##  +  xy. 

Therefore,  just  as  2  +  3  =  5, 

so  2  any  +  3xy  =  (2  +  3)ojy  =  6ajy. 

That  is,  to  add  fo'fce  terms,  add  their  numerical  coefficients  and 
annex  to  the  sum  their  common  literal  part. 

x  Ex.  l.     Add  -  7  ab  to  4  a&. 
We  have    4  a&  +  (-  7  06)  =  [4  +  (—  7)]  a&  =  -  3  a&. 

Ex.  2.     Find  the  sum  of  3  a,  —  5  a,  8  a,  —  4  a. 

Uniting  the  positive  terms  by  themselves,  and  the  negative 
terms  by  themselves,  we  have 


Ex.  3.     Add  ax  to  to. 

Since  the  sum  of  the  coefficients  of  x  is  a  +  6,  we  have 
aa;  -f-  bx  =  (a  +  &)  %• 

EXERCISES  II. 


Add: 

1. 

2. 

3.               4. 

5. 

6. 

a 

-36 

2x         -    3m 

7a 

—  5# 

2a 

-56 

7cc         —  15m 

12  a 

-4ai 

3a 

-26 

5  a?         —  11  m 

-5a 

30 

7. 

8. 

9.               10. 

11. 

12. 

-9a2 
11  a2 
-5a2 

4o?i/ 
—  15  #2/ 
12  xy 

—  7  a?2?/         aa? 
—  6  x*y        ex 

ay 
-by 
cy 

—  mx2 
nx2 
—  px2 

8-9]  ADDITION   AND   SUBTRACTION.  45 

Find  the  sum  of  : 

13.  4a,  5  a,  7  a,  9a.  14.    —5  a,  —  3x,  -9a;,  -13a?. 

15.  6a2,  -3a2,  lla2,  -2a2. 

16.  —  11  xy,  17  xy,  5xy,  —  4  icy. 

17.  8a2&,  -3a26,  27a26,  -Ila2&,  -21a26. 

18.  m  +  n,  —  5  (m  +  n),  9(m  +  n),  —  4  (m  +  n). 

19.  3(a2  +  ^),  -8(a2  +  &),  -14(a2  +  6),  a2  +  6. 

i 

Simplify  the  following  expressions  : 

20.  5  x  —  13  x  +  9  cc.  21.   7  a  —  9  a  —  4  a. 
22.  5m  +  13m-8m.             23.    a2  -  7  a2  -f-  5  a2. 

24.  -  a?b  +  15  a26  -  8  a?b 

25.  2a3- 

26.  -7 

27.  12  a36  -  15  a36  -  8  o86  +  20  a36  - 

28.  a  +  &  -  3  (a  +  &)  4-  8  (a  +  6)  +  5  (a  +  6)  -  10  (a  +  &). 

29.  5(x2  +  y2)  +  8(x24-2/2)-ll(^ 

30.  aj  +  ^aj-|a-Jaj.  31. 

32.  ia  —  ^a  +  fa  —  fa  +  fa  +  fa  —  Jgi-a. 

Simplify  the  following  expressions,  first  removing  paren- 
theses : 

33.  2a  —  [—  4a—  (—  6a)].       34.   m  +[2m  -  (3m  -  4m)]. 

35.  6y-[5y-4y-(-3y  +  2y)]-y. 

36.  x  —  x  —  2x—x  —  3x  —   x  — 


Subtraction  of  Like  Terms. 

9.   Like  Terms  can  be  united  by  subtraction  into  a  single 
like  term. 

Just  as  5-2  =  3, 

so  5  a  —  2  a  =  (5  —  2)  a  =  3  a. 

That  is,  to  subtract  like  terms,  subtract  their  numerical  coeffi- 
cients, and  annex  to  the  remainder  their  common  literal  part. 


46  ALGEBRA.  [Cn.  Ill 

Ex.1.     Subtract  -5x2y  from  -7  tfy. 
We  have 


=  (-  7 

EXERCISES  III. 


Subtract  : 

1. 

2. 

3. 

4. 

5. 

6. 

a 

3x 

—  5m 
2m 

-Sy 

—  3a 

11  x 

-ox 

7. 

8. 

9. 

10. 

11. 

12. 

-la 

-llm 

-12m 

3  a2 
5a2 

7m3 
8m3 

-3ai 

-8JJ 

13.   13  a?b  from  15  a?b.  14.    -Itff  from 

15.   f  ^?/2  from  |  ay2.  16.    —  |  ab5  from  |  a65. 

17.   2  (a  +  6)  from  -  3  (a  +  6).     18.   or2  +  /  from  -  2  (a2  +  2/2). 

Addition  of  Multinomials. 

10.   Unlike  Terms  are  added  by  writing  them  in  succession 
each  preceded  by  the  sign  +  . 

Ex.1.     Add36to2a.     Wehave2a  +  36. 
Ex.  2.     Add  -  3  x2  to  2  y\     We  have 


Such  steps  as  changing   +  (—  Sx2)   into   —  Sx2,  should  be 
performed  mentally. 

11.   A  multinomial  consisting  of  two  or  more  sets  of  like 
terms  can  be  simplified  by  uniting  like  terms. 

Ex.  1.     2a-36-5a      46  =  2a-5a 


12.   If  two  or  more  multinomials  have  common  like  terms, 
these  terms  can  be  united. 


9-12]  ADDITION  AND   SUBTRACTION.  47 

Ex.  l.     Add  -2a  +  36to3a-56. 

We  have  (3a-56)  +  (-2a  +'3  6)  =3a-5b  -2  a  +  36, 

=  a-26. 

In  adding  multinomials,  it  is  often  convenient  to  write  one 
underneath  the  other,  placing  like  terms  in  the  same  column. 

Ex.  2.     Find  the  sum  of  -  4  or2  +  3  /  -  8  z2,  2x2  —  3z2,  and 


We  have  -±x2  +  3y2  -Sz2 

-3z2 


It  is  evidently  immaterial  whether  the  addition  is  performed 
from  left  to  right,  or  from  right  to  left,  since  there  is  no  carry- 
ing as  in  arithmetical  addition. 


EXERCISES  IV. 


Add: 

1. 

2. 

3. 

4. 

5. 

6. 

a 

2 

3 

-4 

a 

—  m 

1 

b 

x 

c 

b 

—  n 

7. 

a  to  a2. 

8. 

—  #  to  x2. 

9. 

—  2m  to  n. 

10. 

x2  to  -2^. 

11. 

/y*yt    "\"O    7/2 

12. 

a2b  to  ab2. 

Simplify  the  following  expressions  by  uniting  like  terms  : 
13.   a  +  2  +  a-  2.  14.   56-3-46  +  4. 

15.    Wx-$  +  5-7x.  16.    9m-3n-8m-f 

17.  _  as  _  5  a2  +  4  a2  +  2  a2  +  2  a3. 

18.  ab  -  3  a262  +  5  ab  -  8  a262  +  4  06  +  a262. 

19.  -  3(a2  +  6)+  5  (a  +  62)+  4(a2  +  6)-  4  (a  +  b2). 


Simplify  the  following   expressions,  first  removing  paren- 
theses : 

20.   a  +  1  —  (2  —  3  a).  21.   5x  —  (—  2y  +  3  a?). 

22.   «  —  2—2    +  3a;  —  3a  —  4. 


48  ALGEBRA.  [Cn.  Ill 


23.  2w  +  3w  —  (5m  —  4n)  —  (— 

24.  2a#  +  5y«  -(2ajy  -  3yz)  +  2ajy  ~(3ajy  -  2  ys)  +  5  2/2. 

25.  a  —  [3a-  (2a  +  6)]—  (36  —  5). 

26.  3a—  [>  +  3y—  (y  —  2a)]. 

27.  8m  —  [m—  (3m  —  ri)  +  (2m  —  3w)]. 

Find  the  values  of  the  expressions  in  Exx.  20-24, 

28.  When  a  =  1,  a;  =  3,  y  =  —  5,  z  =  10,  m  =  4,  n  =  —  7. 

29.  When  a  =  -  3,  a;  =  6,  y  =  —  7,  z  =  8,  w  =  —  1,  w  =  —  2. 

Find  the  sum  of  the  following  expressions  : 

30.  5  a  +  2  6,  3a  —  46,   —  7  a  +  3  6,  9  a  —  6. 

31.  7x  —  3y,  5a  +  42/,  -10ic  +  4?/,  3x-7y. 

32.  a  +  26-3c,  2a-36  +  c,  2a  +  56  +  2c. 

33.  a2  +  2  a  4-  1,  a2  —  3  a  —  2,  a2  +  4  a  +  2. 

34.  ^-5a  +  6,  3ar2  +  2z-7,  Ga^  +  Saj  +  l. 

35.  2  ab  +  3  oc  —  4,  3  a6  —  5  ac  +  2,  —  5  ab  +  3  ac  +  8. 

36.  a2  -  3  ab  +  62,  2  a'2  +  2  a6  -  62,  db  -  2  a2. 

37.  a3  -  5  a26,  7  a26  -  63,  a3  +  63. 

38.  a3  —  3^  +  58  —  1,  7^  +  2^-6^  +  4, 


39.    x3  +  5  x*y  —  7  xy2  —  2  f/3,      —  2  a?3  +  6  of?/  +  11  ay2  —  15  y^, 


40.    ct4  +  2a2-5a-3,   -  3 a4  +  2 a3  +  6 a  -  4, 

2  a4  -  7  a3  +  3  a2  +  9,  5  a4  -  7  a3  -  5  a2  +  a. 

42.    a?  —  2  (a  +  6)2  +  62,  a2  +  3  (a  +  6)2,   —  a2  —  2  62. 

44.  |a26-ia62,   -ia26  +  i«62,   -|a26- 

45.  l-x2-^ 


12-14]  ADDITION   AND   SUBTRACTION.  49 

Subtraction  of  Multinomials. 

13.  Unlike  Terms  are  subtracted  by  writing  them  in  succes- 
sion, each  preceded  by  the  sign  —  . 

Ex.    Subtract  —llm  from  2  n.     We  have 
2  n  —  (—  11  m)  =  2n  +  llm. 

14.  If  two  multinomials  have  common  like  terms,  these 
terms  can  be  united. 

Ex.  1.   Subtract  —  2  a  +  3  b  from  3  a  —  5  b. 
We  have  (3a-5&)-(-2a 


Ex.2.   Subtract  2x2-6x-3  from  3x2-5x  +  l. 

Changing  mentally  the  signs  of  the  terms  of  the  subtrahend 

and  adding,  we  have 

3x2-5x  +  l 

2x2-6x-3 


Ex.  3.  Subtract  2x2-3z2  from  -4x2  +  3y2,  and  from  the 
result  subtract  2y2  +  5z2. 

When  several  multinomials  are  to  be  subtracted  in  succes- 
sion, the  work  is  simplified  by  writing  them  with  the  signs  of 
the  terms  already  changed.  We  then  have 


2X2  +3z2 

-2y2-5z2 


y2-2z2 
EXERCISES  V. 

Subtract : 

1.                 2.  3.  4.                         5.                       6. 

1                  3  x  x2              —  mn                  a-b 

a             —b  y  —x                     m  —  ab2 


50  ALGEBRA.  [Cn.  Ill 

7.  3  a  —  2  b  from  4  a  —  3b.     8.    —  5  #  -}-  4  ?/  from  —  4  £  -f-  5  ?/. 

9.  7  m  +  2  ?i  from  —  3  m  -f-  3  n. 

10.  2^  —  3  a;  from  3  0^  —  2  a;. 

11.  5a- 

12.  2^ 

13.  2  xy  +  5  cez  —  7  ?/z  from  5  a;?/  +  3  xz  —  6  yz. 

14.  2a2-3a6-1262from3a2-a6-1152. 
15. 

16.  a*- 

17.  2  a3  -  a26  -  b3  from  3  a3  +  2  a26  +  3a&2. 

18.  o*-x-l  from  a3  +  2^. 

19.  2  (a?  +  ?/)  —  5  2  from  3  (a;  +  y)  —  4  2. 

20.  6(a-fe)-3a  +  62from  5(a  —  6)  -  2a  +  a2. 

21.  From  the  sum  of  5ic  —  5^  +  32;  and  4#-f-4;?/  —  22  sub- 
tract  8#-2?/-2z. 

22.  From  a2  —  ab  +  62  subtract  the  sum  of  2  a2  -  3  a&  +  5  b2 
and  a2  +  ab  —  4  62. 

23.  How  much  does  m2  -f-  n2  exceed  m2  —  n2  ? 

24.  How  much  does  1  —  #2  exceed  2  —  3  a?2  ? 

25.  What  expression  must  be  added  to2a  —  36-f-4cto  give 


26.  What  expression  must  be  added  to  xy  +  xz  -\-  yz  to  give 
a-2  4.  2/2  +  3,2  ? 

27.  What  expression  must  be  subtracted  from  a2  +  ab  +  Z>2  to 
give  a2  -  2  a&  +  Ir  ? 

28.  What  expression  must  be  subtracted  from  x2  —  2xy  +  y* 
to  give^  +  2^  +  2/2? 

29.  What  expression  must  be  added  to  x2  -f-  x  +  1  to  give  0  ? 

Ifo?  =  2a-36  +  4c,  y  =  —  3a  +  2b  -7  c,  z  =  9a-7b  +  6c, 
find  the  values  of 

30.  x+y+z.        31.    a;—  y-\-z.       32.    x+y—z.       33.    x—y—z. 


14-17]  PARENTHESES.  f>1 


If  A=$x-ly+$z,  B=-$ 
find  the  values  of 

34.    A  +  B  +  C.  35.    A  -  B  +  C. 

36.   A  +  B  -  C.  37.   A  -  B  -  C. 

PARENTHESES. 

15.  The  use  of  parentheses  has  been  briefly  discussed  in 
Ch.  II.,  Arts.  23-25.     It  is  frequently  necessary  to  employ 
more  than  two  sets  of  parentheses,  and  to  distinguish  them 
the  following  forms  are  used  : 

Parentheses,  (  )  ;  Brackets,  [  ]  ;  Braces,  {  j  . 

A  Vinculum   is  a  line  drawn  over  an  expression,  and  is 
equivalent  to  parentheses  inclosing  it. 

E.g.,  (a  +  b)  (c  —  d)  =  a  +  b  -  c  —  d. 

16.  The  principles  given  in  Ch.  II.,  Arts.  23-24,  are  to  be 
applied  successively  when  several  sets  of  parentheses  are  to  be 
removed  from  a  given  expression. 

17.  In  removing  parentheses  we  may  begin  either  with  the 
inmost  or  with  the  outmost. 

The  following  example  will  illustrate  the  method  of  remov- 
ing parentheses,  beginning  with  the  inmost  : 

Ex.  4  a  -  1  3  a  +  [2  a  -  (a  -  1)]  } 


=  4a  —  4a  —  1=—  1. 

When,  in  such  examples,  we  come  to  one  of  a  pair  of  paren- 
theses, (,  or  [,  or  J,  we  must  look  for  the  other  of  like  form. 
We  then  treat  all  that  is  contained  in  each  pair  as  a  whole. 

EXERCISES   VI. 

Simplify  each  of  the  following  expressions  : 


1.  2x-3y-[5x-(2y-3x-y)]. 

2.  a  +  2&-[6a-{3&-(6a-6&){]. 


52  ALGEBRA.  [Cn.  Ill 

4.   6a-[7a-J8tt-(9a- 


5.  a  -  {5  6  -  [a  -  (3  c  -  3  6)  +  2  c  -  (a  -  2  6  -  c)]  {. 

6.  (7a-6)-{4ar-[2a-l-(3-4a-5)]}. 


7.  a  -  [a?  -  1  2  a:  -  3  -  [4  x  -  5  -  (6  x  -  1  x  -  8)]  \  ]  . 

8.  a-3a-a-b  +  5a-b-7a-6  _8a  —  6 


,  9.    Find  the  values  of  the  expressions  in  Exx.  1-5,  when 
«  =  -  3,  6  =  4,  c  =  —  5,  x  =  8,  y  =  -  9. 


18.  Ex.  l.  Express  &(x  —  y)+y  —  x  as  a  product,  of  which 
one  factor  is  x  —  y. 

We  have  4  (x  —  y)  +  ?/  —  x  =  4  (a?  —  y)  —  (x  —  y)  =  3  (x  —  y). 

The  sign  -f-  °r  —  before  a  pair  of  parentheses  can  evidently 
be  reversed  from  +  to  —  ,  or  from  —  to  +,  if  the  signs  of  the 
terms  within  the  parentheses  be  reversed. 

Ex.  2.   7aj-l-3l-oj=7a?- 


EXERCISES   VII. 

Write  each  of  the  following  expressions  as  a  product,  of 
which  the  expression  within  the  parentheses  is  one  of  the 
factors  : 

1.   3(a—  &)  —  a  +  &.  2. 

3.   3ra-5n-4(5n-3ra).       4.   1  -  an  +  3(an  -  1). 

5.    5(x2  —  x+1)—  x*  +  x—  1.       6.   a  —  t/  —  2  —  6(y  +  z  —  x). 


Write  each  of  the  following  expressions  as  a  single  product, 
of  which  the  expression  within  the  first  parentheses  is  a  factor  . 

7.     2oj-l-3l-2oj.  8.    22m- 


9.   5  (x2  —  y2)  +  2  (y2  —  x2).         10.    7  (xy  —  z)  —  (z  —  xy). 

Simplify  the  following  expressions    without   removing   the 
parentheses : 

11.    (a  —  &)  c  +  (6  —  a)  c.  12.    5  (x  —  y)  z  -f  5  (y  —  x)  z. 


17-21]  EQUATIONS  AND   PROBLEMS.  53 

EQUATIONS   AND   PROBLEMS. 

19.   Ex.     Find  the  value  of  x  from  2x  —  5  =  T  +  x. 
Adding  5  to  both  members  of  the  equation,  we  obtain 

2x  —  5  +  5  = 
or,  since  —5  +  5  =  0,         2#  = 


Subtracting  x  from  both  members  of  the  last  equation,  we 

have 

2x  —  x  =  7  +  5+iP—  a; 

or,  since  x  —  x  =  0,      2  x  —  x  =  1  +  5.  (1) 

Uniting  terms,  x  =  12. 

Check:  2x12-5  =  7  +  12,  or  24-5  =  7  +  12,  or  19  =  19. 

20.  Observe  that  equation   (1),  Art.  19,  could   have  been 
obtained  directly  from  the  given  equation  by  transferring  the 
term  —  5,  with  sign  changed,  to  the  second  member,  and  the 
term  +  aj,  with  sign  changed,  to  the  first  member. 

That  is,  any  term  may  be  transferred  from  one  member  of  an 
equation  to  the  other,  if  its  sign  be  reversed  from  +  to  —  ,  or 
from  —  to  +. 

21.  Ex.   Find  the  value  of  x  from  the  equation  x—  3=8—  3. 
Adding  3  to  both  members  of  the  equation,  we  obtain  : 

x  _  3  +  3  =  8  -  3  +  3  ; 
or,  since  —  3  +  3  =  0,  x  =  S. 

Check:  8  -  3  =  8  -  3,  or  5  =  5. 

Observe  that  this  step  is  equivalent  to  dropping  the  common 
term  —  3  from  both  members. 

That  is,  the  same  term,  or  equal  terms,  may  be  dropped  from 
both  members  of  an  equation. 

This  step  is  called  cancellation  of  equal  terms. 


T>4  ALGEBRA.  [Cn.  Ill 

22.  These  examples  illustrate  the  following  method : 
Transfer  all  the  terms  containing  the  unknown  number  to  one 

member  of  the  equation,  usually  to  the  first  member,  and  all  the 
terms  containing  known  numbers  to  the  other  member. 

Unite  like  terms. 

Divide  both  members  by  the  coefficient  of  the  unknown  number. 

Check  by  substituting  the  value   thus   obtained  in   the   given 
equation. 

23.  Pr.  A  boy  being  asked  his  age,  replied,  "  If  10  is  added 
to  twice  the  number  of  years  in  my  age  the  sum  will  be  40." 
How  old  was  the  boy  ? 

Let  x  stand  for  the  number  of  years  in  his  age. 
Then  2  x  stands  for  twice  that  number  of  years. 
The  problem  states, 

in  verbal  language  :  twice  the  number  of  years  in  the  boy's  age 
plus  10  is  equal  to  40 ; 

in  algebraic  language  :         2  x  +  10  =  40. 
Transferring  10,  2  x  =  30. 

Dividing  by  2,  x  =  15. 

The  boy  was  15  years  old. 
Check:  2  x  15  +  10  =  40,  or  30  + 10  =  40,  or  40  =  40. 

EXERCISES   VIII. 

Solve  each  of  the  following  equations  : 

1.  #  +  4  =  9.  2.   3  +  .T  =  10.  3.   x  —  5  =  6. 

4.  15-a  =  10.  5.   ll-x  =  13.          6.   3^  +  2  =  11. 

7.  5x-3  =  17.  8.    7 +  12  a;  =  31.        9.    41 -17  a;  =7. 

10.  15  =  3  +  4#.        11.   19  =  13-6x.      12.   14  =  8-3o;. 

13.  9  +  5  x  =  13  +  4  x.  14.    8  x  -  5  =  10  x  -  11.  • 

15.  18-5x  =  33-8x.  16.    14  a?- 13  =  7  a? +  29. 

17.  3z-4  +  5o;  =  7a;  +  9.         18.    4a;-9  =  8a;-3-2aj. 


22-23]  EQUATIONS   AND   PROBLEMS.  55 


19.  2 

20.  11  #  —  15  —  4  x  =  2x  +  5  —  5  x. 

21.  13^-25  +  7  x  =  87  +  9  x  +  9. 

22.  5«  +  14-8x  =  3x-16-4ic. 

23.  6  x  +  7  -  15  x  +  23  =  36  x  +  15. 

24.  6  a;  -  25  +  3  x  -  14  x  =  25  -  3  a. 

25.   4  a  +  (2  a  —  3)  =  15.  26.   2  x-  (5  <c  +  5)  =  7. 

27.   5a?—  (3a  —  7)  =  17.  28.    7a—  (3a;  —  11)  =  4. 

29.    14a?-{3aj-(2-a;)}=22.     30.   3  a;-  7  -  (5  a?  +  17)=0. 

31.  6  a?  -  [7  »-  (8  a  -18)]=  16. 

32.  6-J5-(4-S3-[2-(l-^)]OI  =  4. 

33.  If  19  is  added  to  a  number,  the  sum  will  be  40.     What 
is  the  number  ? 

34.  A  man  invests  $2100.     How  much  must  he  gain  to 
have  $3600? 

35.  What  number  increased  by  43  gives  its  double  ? 

36.  What  number  is  16  less  than  three  times  itself  ? 

37.  A  pole  34  feet  long  is  divided  into  two  parts,  so  that 
one  part  is  8  feet  shorter  than  the  other.     What  is  the  length 
of  each  part  ? 

38.  In  a  number  of  2  digits,  the  tens'  digit  exceeds  the  units' 
digit  by  3.     If  the  sum  of  the  digits  is  13,  what  is  the  units' 
digit  ?     What  is  the  number  ? 

39.  What  is  the  number  next  greater  than  8  ?     Next  less  ? 
Next  greater  than  x  ?     Next  less  ?     Next  greater  than  x  +  4  ? 

40.  The  sum  of  two  consecutive  numbers  is  31.     What  are 
the  numbers  ? 

41.  The  sum  of  three  consecutive  numbers  is  24.     What  are 
the  numbers  ? 

42.  The  difference  between  two  numbers  is  7,  and  the  smaller 
number  is  9.     What  is  the  greater  number?     If  the  greater 
number  is  x,  what  is  the  smaller  number  ? 


56  ALGEBRA.  [Cn.  Ill 

43.  The  difference  between  two  numbers  is  8,  and  their  sum 
is  38.     What  are  the  numbers  ? 

44.  The  difference  between  two  numbers  is  3,  and  their  sum 
is  equal  to  nine  times  their  difference.    What  are  the  numbers  ? 

45.  A  father  is  40  years  older  than  his  son.     If  six  times 
the  son's  age  is  equal  to  the  sum  of  their  ages,  how  old  is 
each? 

46.  The  length  of  a  room  is  three  times  the  breadth.    If  the 
length  is  20  feet  more  than  the  breadth,  what  are  the  dimen- 
sions of  the  room  ? 

47.  A  man,  being  asked  the  time,  replied,  "If  18  is  sub- 
tracted from  four  times  the  hour  it  now  is,  the  remainder  will' 
be  the  hour."     What  was  the  hour  ? 

48.  Three  times  a  number  exceeds  12  by  as  much  as  12 
exceeds  the  number.     What  is  the  number  ? 

49.  A  has  $  125  and  B  has  $  45.     How  many  dollars  must 
A  give  'B  in  order  that  they  may  have  equal  amounts  ? 

50.  A  pile  stands  3  feet  above  the  water.     If  1  is  in  the 
water  and  1  in  the  earth,  how  long  is  the  pile  ? 

51.  Two  vessels  together  hold  9  gallons.      If  the  smaller, 
when  empty,  is  filled  from  the  larger,  when  full,  there  will 
remain  3  gallons  in  the  latter.     How  many  gallons  does  each 
vessel  hold  ? 

52.  Three  boys,  A,  B,  and  C,  pull  100  pounds.     A  pulls  20 
pounds  more  than  B,  and  B  pulls  8  pounds  less  than  C.     How 
many  pounds  does  each  boy  pull  ? 

53.  A  pole  is  divided  into  three  parts.     The  second  is  three 
times  as  long  as  the  first,  and  the  third  is  6  feet  longer  than 
the  first.     The  length  of  the  pole  is  equal  to  the  excess  of  60 
feet  over  the  smallest  part.      What  are  the  lengths  of  the 
parts,  and  the  length  of  the  pole  ? 

54.  In  a  number  of  two  digits,  the  units'  digit  is  three  times 
the  tens'  digit.     The  number  is  equal  to  8  more  than  three 
times  the  units'  digit.     What  is  the  number  ? 


23-25]  MULTIPLICATION.  57 

MULTIPLICATION. 
Product  of  Powers. 

24.   Ex.  1.    a3  x  a4  =  (aaa)  (aaaa)  =  aaaaaaa  =  a7  =  a3+4. 
Ex.  2.   xx2x3  =  x  (xx)  (xxx)  =  xxxxxx  =  XQ  =  xl+2+s. 
These  examples  illustrate  the  following  principle  : 

The  exponent  of  the  product  of  two  or  more  powers  of  the  same 
base  is  the  sum  of  the  given  exponents;  or  stated  symbolically, 

ama"  =  am+n  ;  amana^  =  am+n+P  ;  etc. 


EXERCISES   IX. 

Express  each  of  the  following  products  as  a  single  power  : 

l.  32x3.  2.  53x52.          3.  64x63.  4.  (-5)452. 

5.  (_6)S(-6)4.       6.  25(-2)7.       7.   83(-8)4.  8.  (-7/T3. 

9.  X*XXG.  10.  (-y)V.     11.  (-«)3(-«)4.     12.  (-x^x3. 

13.  a3a5a7.  14.   x\—  x)6x*.  15. 

16.  (xy)s(xy)4.  17.    (2  a&)3[- 

18.  (a  +  &)3(a  +  &)5-  19-    [_-(x-y)-\*(x- 

20.  xnx^.  21.  ana2.  22.  xn~lx.  23. 

24.  2n+1i8B-1.  25.  ic2n-Vn+3.     26.    bm+1bn~l.  27. 


Degree.     Homogeneous  Expressions. 

25.   An  integral  term  which  is  the  product  of  n  letters  is 
said  to  be  of  the  nth  degree. 

Thus,  the  Degree  of  an  Integral  Term  is  indicated  by  the  sum 
of  the  exponents  of  its  literal  factors. 

E.g.,  3  ab  is  of  the  second  degree  ;  2  x2?/,  =  2  xxy,  is  of  the 
third  degree. 

The  Degree  of  a  Multinomial  is  the  degree  of  that  term  which 
is  of  highest  degree. 

E.g.,  the  degree  of  x2y  -f-  xy3  —  x2y*z  is  the  degree  of 
t.e.,  the  sixth. 


58  ALGEBRA.  [Cn.  HI 

26.  It  is  often  desirable  to  speak  of  the  degree  of  a  term, 
or  of  an  expression,  in  regard  to  one  or  more  of  its  literal 
factors. 

E.g.,  the  term  ax\f  is  of  the  fifth  degree  in  x  and  y,  of  the 
first  degree  in  a,  of  the  second  degree  in  x,  of  the  third  degree- 
in  y,  etc. 

The  expression  ax2  +  2  bxy  +  cy2  is  of  the  second  degree  in 
x,  in  y,  and  in  x  and  y. 

27.  A  Homogeneous  Expression  in  one  or  more  letters  is  an 
expression  all  of  whose  terms  are  of  the  same  degree  in  these 
letters. 

E.g.,         a2  +  2  ab  -f-  b2  is  homogeneous  in  a  and  b. 

28.  If  the  terms  of  a  multinomial  be  arranged  so  that  the 
exponents  of  some  one  letter  increase,  or  decrease,  from  term 
to  term,  the  multinomial  is  said  to  be  arranged  to  ascending, 
or  descending,  powers  of  that  letter.     The  letter  is  called  the 
letter  of  arrangement. 

E.g.,  x4  —  3  x3y2  -f-  2  x2y  +  xif  is  arranged  to  descending  powers 
of  x,  which  is  then  the  letter  of  arrangement ;  or,  when  written 
x*  4-  2  x2y  —  3  x*y2  -f-  xy*,  to  ascending  powers  of  y,  which  is  then 
the  letter  of  arrangement. 

EXERCISES  X. 

What  is  the  degree  of  2  cptfahf 
1.   In  a  ?     2.    In  x  ?     3.    In  b  and  y  ?     4.    In  a,  b,  x,  and  y  ? 

What  is  the  degree  of  the  expression  a3x*— 6  <j&b*x*y+5  abx2y2 
5.    In  x  ?  6.    In  y?  7.    In  a  ?  8.    In  b  ? 

9.    Arrange  2  x  —  3  x5  -(-  7  —  2  x4  -f-  3  x2  to  ascending  powers 
of  x ;  to  descending  powers  of  x. 

10.  Arrange  3  y  —  7  xy8  -f-  o  x8y2  +  4  x2y4  to  ascending  powers 
of  x  -,  to  ascending  powers  of  y. 


26-29]  MULTIPLICATION.  5 

Multiplication  of  Monomials  by  Monomials. 
29.    Ex.  1.  3  a  x  5  b  =  3  x  5  x  a  x  b, 

=  15  db. 

Ex.  2.          2  x  x  (-  4  /)  =  2  (-  4)  a*/2  =  -  8  a/. 
Ex.  3.  fa3  x  6  a62  x  11  65  =  f  x  6  x  11  x  a?ab2b5  =  44  a467. 
Ex.  4.  -  3  xm  x  4  x2  =  —  3  x  4  x  ^w^  =  -  12  ^w+2. 

Ex.  5.   5  af+1  x  7  a?"-^  5  x  7  x  ie^V-1  =  35  ^1+1+M-1  =  35  x2n. 


These  examples  illustrate  the  following  method  of  multiply- 
ing two  or  more  monomials. 

Multiply  the  product  of  the  numerical  coefficients  by  the  product 
of  the  literal  factors. 

EXERCISES   XI. 

Multiply  : 


1. 

2. 

3. 

4. 

5. 

6. 

2a 
3 

3x 

—  2 

6 

—  6m 

-8 

10. 

-26 

-  5  w3 

7  >t 

7. 

8. 

9. 

11. 

12. 

-65J 

-«* 
-^ 

7  a2 

:525 

2  abc2 

5  a;-?/^3 
—  3xy~z2 

13.    2(a  +  Z>)  by  3(a  +  ^)2.       14.    -  5 (a  -  yf  by 
Simplify  the  following  continued  products 
15.    3  ab  x  5  be  x  6  ac.  16.    -  7  afy  X  (-  2  ?/2^)  x  3  xz2. 

17.    —xy2x7  bx-z  x  2  6^2.    18.    x2?/^1  x  5  xmy*"  x  (  — 

Multiply : 

19.  2a362c,   -3«&2c3,  a*&4c,   —  5a6c4. 

20.  am+2,  a2"*,  a3~TO,  aTO~n,  a2""81". 

21.  aj"+1,   —  5  a?"-1,  2  x-~m,  xm+2. 


60  ALGEBRA.  [Cn.  Ill 

Multiplication  of  a  Multinomial  by  a  Monomial. 

30.  If  the  indicated  operation  within  the  parentheses  in  the- 
product,  4  (2  -f-  3  —  1),  be  first  performed,  we  have 

4(2+3-l)  =  4x4  =  16. 

But  if  each  term  within  the  parentheses  be  multiplied  by  4 
and  the  resulting  products  be  then  added,  we  have 

4x2  +  4x3-4x1  =  8  +  12  -4  =  16,  as  above. 

Therefore    4(2  +  3-1)  =  4x2  +  4x3-4x1. 

This  example  illustrates  the  following  method  of  multiplying 
a  multinomial  by  a  monomial  : 

Multiply  each  term  of  the  multinomial  by  the  monomial,  and 
add  algebraically  the  resulting  products.  That  is, 

a  (b  +  c  —  (/)  =  ab  +  ac  —  ad. 

This  principle  is  called  the  Distributive  Law  for  multiplica- 
tion. 

31.  Ex.  1.    Multiply  (x  -  y)  by  3. 

We  have  3(x  —  y)  =  3x  —  3y. 

Ex.  2.    Multiply  3x  —  2y  —  7z  by  —  4  #. 
We  have 


=  -  12  x2  +  8  xy  +  28  xz. 

Such  steps  as  changing  (—4  a;)  (3  a;)  into  —  12  #2,  —  (— 
into  +8xy,  and  —(—4  x)  (7  z)  into  +28x2!,  should  be  per- 
formed mentally. 

The  work  may  be  arranged  as  in  arithmetic,  by  placing  the 
multiplier  under  the  multiplicand  : 

3x-2y-7z 


-  12  x'2  +  8  xy  +  28  xz 

It  is  customary  to  multiply  from  left  to  right,  instead  of 
from  right  to  left  as  in  arithmetic. 


30-31]  MULTIPLICATION.  61 

EXERCISES   XII. 

Multiply : 

1.  x  +  1  by  3.  2.    a -3  by  5.  3.    2m +  5  by  —3. 

4.  3x-7by-8.     5.   2a  +  36by3a,     6.   5  x  —  3y  by  2 x. 

7.  6a2-56  by  56.  8.  3x-5y2  by  -6xy. 

9.  Stf  +  Sy2  by  2xy.  10.  a  +  6  -  c  by  5. 

11.  a-^-z  by  -3.  12.  3a  +  26-5c  by  4. 

13.  5ra  — 3w— -4_p  by  —3.          14.  2a  — 76  +  3cby  —  5  a. 

15.  —  5  x-  +  3  ?/2  —  2  z2  by  —  2  cc^. 

Multiply  a2  -  3  a  +  1  by 

16.  2  a.  17.   -36.  18.   5a6.  19.    —  6a263. 

Multiply  x*y  +  3  xy  —  5  ?/2  by 
20.    -3s*         21.    -5i/2.  22.    ~6x*y.         23.   Sx2?/2. 

Simplify   the   result   of   substituting  a  +  6  —  c  for  x,  and 
a  —  6  +  c  for  y,  in  the  following  expressions : 

24.   bbx-lay.          25.   3arbx-Uab2y.          26.   7abx  +  2bcy. 

Find  the  values  of  the  results  of  Exx.  24-26 

27.  When  a  =  -  2,  6  =  3,  c  =  -  4. 

28.  When  a  =  5,  6  =  —  7,  c  =  —  5J. 

Multiply  5  xn  -  3  xn~3y2  -f  4  xn~y  +  2/n~4  by 
29.    x3.  30.    -5x2y.  31.   3a?Y-  32-    - 

Simplify  the  following  expressions : 

33.  4o?-2}[a?-3(2-aj)]aj-4}. 

34.  13a~13{10[7(4a-3)-6]-9a}. 

35.  -206-2J«-5[3-2x-6(4ic-7)]-3(5-2a;)|. 

36.  x  +    *x-2-lx-x2 


'2  ALGEBRA.  [Cn.  Til 

Multiplication  of  Multinomials  by  Multinomial? 
32.   Ex.    Multiply  7-5  by  2  +  3. 
If  we  let  a  stand  for  7  —  5,  we  have 


Now  replacing  a  by  7  —  5,  we  obtain 

(2  +  3)  (7  -  5)  =  2  (7  -  5)  +  3  (7  -  5) 

=  2x7-2x5  +  3x7-3x5. 

This  example  illustrates  the  following  method  of  multiply- 
ing a  multinomial  by  a  multinomial  : 

Multiply  each  term  of  the  multiplicand  by  each  term  of  the  mul- 
tiplier, and  add  algebraically  the  resulting  products. 
In  general, 

(a  +  b)  (c  +  d  -  e)  =  a  (c  +  d  -  e)  +  b  (c  +  d  -  e) 
=  ac  +  ad  —  ae  +  be  +  bd  —  be. 

33.  1.   Multiply  -3a  +  26by  2a-3b. 

We  have  -3a  +    26 

2a  -    36 

4  aft 
9a6-668 


-  6  a2  +  13  «6  -  6  62 

The  work  is  arranged  as  follows  :  Write  the  multiplier  under 
the  multiplicand;  the  first  partial  product,  i.e.,  the  product  of  the 
multiplicand  by  the  first  term  of  the  multiplier,  under  the  multi- 
plier ;  the  second  partial  product  under  the  first;  and  so  on, 
placing  like  terms  of  the  partial  products  in  the  same  column. 

Ex.  2.   Multiply  x  +  a  by  x  +  6. 

We  have  x  +  a 

x  +  b 
x~  +  ax 

ab 


4.  (a  4.  ft)  a-  4. 


-33]  MULTIPLICATION. 

Ex.  3.    Multiply  4a2  +  l-2a-8a3  by  1+2  a. 
Arranging  to  ascending  powers  of  a,  we  have 


2a-4a2+8a3-16a* 
1  -  16  a4 

Ex.  4.    Multiply  tf  +  yZ  +  l  —  xy  —  x  —  y  by  x 
Arranging  to  descending  powers  of  x,  we  have 

x2  —  xy  —  x  +  /  —  y  -f  1 
a?  +   y  +  1 


a?3  —  x2?/  —  3?  -\-  My2  —    xy  -\-x 

x2y         —xy*-     xy 
_  +  ^  —    xy  —  x         +  /  —  y  +  1 

a?  -Sajjy        +^  +1 


Ex.  5.    Multiply  2  am+1  —  5  of1  +  7  x7"-1  by  x2m  —  x*m~\ 
We  have 

—  5  xm  -f-    7  x"-1 


2  ^'"+1  -  7  or1*  +  12  or3—1  -  7 

EXERCISES  XIII. 

Multiply : 

l.  a  +  1  by  a  +  2.  2.  a  +  1  by  a  -  2. 

3.  m  —  5  by  m  +  3.  4.  i/ —  6  by  T/  — 5. 

5.  m  -  12  by  m  -  3.  6.  a  -  12  by  a  -  15. 

7.  2  a;  +  1  by  x  +  3.  8.  3  a  +  5  by  2  a  -  3. 

9.  llm  — 6  by  2m  — 5.  10.  15  05  —  8  by  10  a  — 3. 

11.  x  +  y  by  x  —  y.  12.  2  a  +  b  by  3  a  —  b. 


64  ALGEBRA.  [Cn.  Ill 

13.    3m—  2  n  by  5  m  +  3  n.         14.    5x  —  6y  by  3  x  —  2  y. 
15.    2#2  +  7y  by  5x2-3y.          16.    11  m2  +  On  by  5m2  -  7  n. 
17.    2a2  +  368  by  4a2-562.        18.    3ar  +  ±xy  by  2x2  +  3xy. 
19.    7a2  + 2a&  by  3a2-5ab.      20.    6a2-5a?/  by  3x2-2xy. 
21.   x2  +  x  +  l  by  x  —  1.  22.   0^  —  0;  + 1  by  av+ 1. 

23.   a2  +  5a-6  by  a  -  3.  24.   x2-llx  +  l2  by  a? -8. 

25.    2a2  +  3a-o  by  3a-2.        26.    6aj2-7aj  +  2  by  6aj-7. 
27.   5  x2  -  2  a?  +  1  by  5  x  +  2.       28.   3  x2  +  4  aj  -  5  by  3  x  -  4. 
29.   x2  +  2xy  +  y2  by  a-  +  ?/.          30.   a2  —  2  ab  +  62  by  a  —  &. 
31.   Xs  —  x2  —  oj  +  1  by  aj  +  1.        32.   x3  +  x2H-ic+l  by  a;  — 1. 

33.  8a?  — 4a^  +  2o;  — 1  by  2a?  +  l. 

34.  cc3  +  a?2?/  +  ^2/2  +  y3  by  x  —  y. 

35.  27a3  +  18a26  +  12a&2  +  8&3  by  3a-26. 

36.  2a  +  36  +  ocby2aH-36  — 5c. 

37.  Ga^  +  Sa  +  l  by  Ga^- 

38.  l  +  ^  +  ^y  by  1-ajy- 

39.  2a2-3a6  +  562  by  2a2 

40.  2  «2  +  3  xy  +  4  /  by  3  x2  -  4  xy  +  ?/2. 

41.  0^-2^  +  3^-1  by  x2-3x  +  2. 

42.  a54  — 5i»2  +  6.T  — 3  by  x2  +  5x  —  4. 

43.  x4-6£c3  +  2x  +  5  by  3^-2^  +  5. 

44.  x3  —  4  or?/  4-  2  a#2  —  t/3  by  x2  —  3  xy  +  y2. 

45.  x4  +  2^  +  a?2-4ar-li  by  ^-2^4-3. 

46.  x2  —  xy  4-  y2  -f  #  +  ?/  +  1  by  #  4-  y  —  1. 

47.  xn  —  2  a**-1  —  3  xn~2  —  5  ccn~3  by  x  +  1. 

48.  5  a;"  +  3  a;"-1  —  8  a;n~2  —  3  by  aj  — 2. 

49.  an+1-5an-f  7an~1-3  by  a2  +  a  +  l. 

50.  aj3"  —  o;2n  +  xn  —  1  by  a;n  + 1. 

51.  a2n  -  2  anln  +  &2"  by  a2"  +  2  a"6n  +  ft2". 


33-34]  MULTIPLICATION.  65 

52.  (x  +  m)(x  +  n).  53.    (x  —  m)  (x  —  ri). 

54.  (x  +  m)(x  —  ri).  55.    (x  —  m)  (x  +  n). 

56.  [>2  -  (a  4-  6)  a;  +  06]  (a  -  c). 

57.  x2  +  a-&a 


Simplify  each  of  the  following  expressions  : 

58. 

59. 

60.     X- 

61. 

62. 

63. 

64.  ^(y  -  2)  +  f(z  -  x)  +  Z20  -  y)  +  (a?  -  y)  (y  -«)(«-  a?). 

65.  (x-^+^-^  +  ^-^ 

66.  (2m2  +  3m-2)(m-l)(2m 

67.  (a? 

68.  (a? 

69.  (a2-a  +  l)(a2  +  a  +  l)(a4-a2  +  l). 

Simplify  each  of  the  following  expressions  : 

70.  (Ja2  +  i&2)(ia2  +  i&2).         71.  Qa2 

72.  (fa-f&  +  fc)(f«-f&  +  fc). 

73.  (Ja;_}y-|-52!)(4*-3y-i2!). 

74.  (21  -3-^  +  31  x2)  (i  x2  +  21  x  +  It). 

75.  (3a2-f-i62-ic2)(fa2-i62--|c2). 

76.  (±ax  +  i  bx2  +  icx3)  (t  aaj  +  t  &*  - 

Zero  in  Multiplication. 

34.    Since        N-  0  =  N(b  —  b),  by  definition  of  0, 

=  Nb  -  Nb  =  0, 
we  have  N  -  0  =  0  and  0  -  M  =  0. 

That  is,  a  product  is  0  if  one  of  its  factors  be  zero. 


66  ALGEBRA.  [CH.  Ill 

EXERCISES   XIV. 

1.  What  is  the  value  of  2 (a  —  &),  when  b  =  a? 

2.  What  is  the  value  of  (a  +  6) (c  —  d),  when  c  =  d? 

3.  What  is  the  value  of  (b  -\-  c)  (a  -+-  b  —  c),  when  c  =  a  -\-b? 

4.  What  is  the  value  of  (a?2  —  9)  (x4  —  7  a?  +  2  x  —  9),  when 
x  =  3? 

For  what  values  of  x  does  each  of  the  following  expressions 
reduce  to  0 : 

6.    (x-£)(x  +  7)?         7.    (x-l)(x-d)? 
25)?         9.   x(x-a)(x-b)(x-c)? 

Equations  and  Problems. 
35.  Ex.   Find  the  value  of  x  from  the  equation 

3(a;-4)  +  5  =  4(>  -3). 

Removing  parentheses,  3x  —  12  -f-  5  =  4  a?  — 12. 
Cancelling  -12,  3z  +  5  =  4#. 

Transferring  terms,  3#  —  4#  =  —  5, 

or  —  x  =  —  5. 

Dividing  by  —  1,  #  =  5. 

Check:  3(5 -4) +5  =  4(5  -  3),  or  3  +  5  =  4x2,  or  8  =  8. 

To  solve  such  equations :  Remove  parentheses,  and  proceed  as 
in  Art.  22. 

Pr.  A  number  of  persons  were  to  raise  a  fund  by  paying  $  5 
each.  Had  there  been  4  persons  more,  each  would  have  had 
to  contribute  only  $  3.  How  many  persons  were  there  ? 

Let  x  stand  for  the  number  of  persons. 

Then  the  number  of  dollars  contributed  was  5  x. 

Had  there  been  4  persons  more,  there  would  have  been 
x  +  4  persons. 

Then  the  number  of  dollars  contributed  would  have  been 
3  (x  +  4). 


34-35]  MULTIPLICATION.  67 

The  problem  implies, 

in  verbal  language  :  the  number  of  dollars  contributed  in  the  one 
case  is  equal  to  the  number  of  dollars  contributed  in  the  other; 

in  algebraic  language  :  5  x  =  3  (x  -f  4). 

Kemoving  parentheses,  5  x  =  3  x  +  12. 

Transferring  3  a?,  2x  =  12. 
Dividing  by  2,  x  =  6. 

Check  :    6   persons   contributed    6  x  5,  =  30  dollars  ;    6  +  4, 
or  10,  persons  would  have  contributed  10  x  3,  =  30  dollars. 

EXERCISES  XV. 

Solve  the  following  equations  : 
1.   6(»  +  l)  =  6.  2.   4  (2.3-1)  =  5. 

3.   3(z  +  5)  +  17  =  26.  4.  14  +  3(7-2z)  =  29. 

5.   15  +  4(8-2z)  =  7.  6.   25-3(5-4a)  =  22. 

7.   27  +  4(2z  -8)  =  12.  8.   11(2-  5  a?)  =47  -30  a. 

9.   12(4»  —  5)  =  13  —  98a?.       10.   7x-  6(10  —  0?)  =  33». 

11.  4(2a5  +  3)  _3(2a?  +  4)  =  10. 

12.  5(3x  +  4)-2(4z-3)  =  54. 

13.  7(2«-3)-ll(5aj-4)  =  64. 

14.  (X-3)(x-4:)=xz  +  5. 

15.  (a;  -  4)  (a;  —  6)=  a;  (a;  —  9). 

16.  («  +  !)(»  +  2)  =  (a;  -3)  (a;  -4). 
17. 


18.  The  sum  of  two  numbers  is  50.     If  five  times  the  less 
exceeds  three  times  the  greater  by  10,  what  are  the  numbers  ? 

19.  Two  boys,  A  and  B,  had  the  same  number  of  apples. 
A  said  to  B  :    "  Give  me  5  apples  and  I  will  have  twice  as 
many  as  you  will  have  left."     How  many  apples  had  each  ? 


68  ALGEBRA.  [C'n.  Ill 

20.  Add  10  to  a  certain  number,   and   multiply  the   sum 
by  2,  or  subtract  8  from  the  same  number,  and  multiply  the 
difference   by   5.      The   results  will  be  equal.     What  is  the 
number  ? 

21.  A  is  30  years  old,  and  B  is  12  years  old.     After  how 
many  years  will  A  be  twice  as  old  as  B? 

22.  A  father  is  30  years  older  than  his  son ;   5  years  ago 
he  was  four  times  as  old.     What  are  the  ages  of  father  and  son  ? 

23.  A  and  B  invested  equal  amounts.     A  gained  $  200,  and 
B  gained  $2600.     If  B  then  had  three  times  as  much  as  A, 
how  much  did  each  invest  ? 

24.  Three  boys,  A,  B,  and  C,  catch  128  fish.     If  B  catches 
10  more  fish  than  A,  and  C  catches  three  times  as  many  as  A 
and  B  together,  how  many  fish  does  each  boy  catch. 

25.  In  one  room  there  are  twice  as  many  persons  as  in  a 
second  room.     If  10  persons  pass  from  the  first  room  into  the 
second,  there  will  be  three  times  as  many  persons  in  the  second 
as  in  the  first.     How  many  persons  are  there  in  each  room  ? 

26.  A  woman  has  enough  money  to  buy  11  yards  of  cloth 
of  one  kind,  or  8  yards  of  another  kind.     If  the  latter  costs 
30  cents  more  a  yard  than  the  former,  how  much  does  a  yard 
of  each  kind  cost  ? 

27.  In  a  stairway  there  are  45  steps  of  a  certain  height.     If 
the  steps  had  been  made  1  inch  higher,  there  would  have 
been  only  40.     How  high  are  the  steps  ? 

28.  The  capacity  of  a  certain  vessel  is  90  gallons.     One 
pipe  lets  in  2  gallons  a  minute  and  a  second  pipe  1  gallon.     If 
the  first  pipe  is  opened  15  minutes  before  the  second,  how  long 
after  the  first  pipe  is  opened  will  the  vessel  be  filled  ? 

29.  A  farmer  has  two  fields  containing  together  5  acres.     A 
offers  to  pay  $  62  an  acre  for  the  first  field  and  $  72  an  acre 
for  the  second.     B  offers  to  pay  $  60  an  acre  for  the  first  field 
and  $  75  an  acre  for  the  second.     If  both  offers  amount  to  the 
same,  how  many  acres  are  there  in  each  field  ? 


35-37]  DIVISION.  69 

30.  The  capacity  of  a  certain  cistern  is  2200  gallons.     One 
pipe  lets  in  80  gallons  in  a  minute,  and  a  second  pipe  50 
gallons.     How  many  minutes  must  the  first  pipe  be  opened 
before  the  second  in  order  that  the  cistern  may  be  filled  4 
minutes  after  the  second  pipe  is  opened? 

31.  One  cask  contains  70  gallons,  and  another  50  gallons. 
If  three  times  as  many  gallons  are  drawn  from  the  larger  as 
from  the  smaller,  the  contents  of  the  smaller  will  be  equal  to 
three  times  the  contents  of  the  larger.      How  many  gallons 
are  drawn  from  each  cask  ? 

32.  A  man  has  $  115  in  two-dollar  bills  and  five-dollar  bills. 
If  he  has  35  bills  altogether,  how  many  of  each  kind  has  he  ? 

33.  A  rides  his  bicycle  12  miles  an  hour,  and  B  his  10  miles 
an  hour.     A  rides  a  certain  number  of  hours,  and  B  rides  2 
hours  longer.     If  they  ride  the  same  distance,  liow  many  hours 
does  each  ride  ? 

34.  Twenty-five  men  were  to  raise  a  certain  fund  by  con- 
tributing equal  amounts.     But  5  men  failed  to  contribute,  and 
in  consequence  each  of  the  remaining  men  had  to  contribute 
$  2  more.     What  was  to  be  the  original  contribution  of  each  ? 
What  was  the  amount  of  the  fund  ? 

DIVISION. 

36.  One  power  is  said  to  be  higher  or  lower  than  another 
according  as  its  exponent  is  greater  or  less  than  the  exponent 
of  the  other. 

E.g.,  a4  is  a  higher  power  than  a3  or  62,  but  is  a  lower  power 
than  a6  or  b7. 

Quotient  of  Powers  of  One  and  the  Same  Base. 

37.  Ex.  a7  -r-  a3  =  (aaaaaad)  -r-  (aaa). 

=  (aaaa)  x  (aota)  -f-  (aaa) 
=  aaaa  =  a4  =  a7"3. 


70  ALGEBRA.  [Cn.  Ill 

This  example  illustrates  the  following  method  of  dividing  a 
higher  power  by  a  lower  power  of  the  same  base : 

The  exponent  of  the  quotient  is  the  exponent  of  the  dividend 
minus  the  exponent  of  the  divisor;  or,  stated  symbolically, 

We  also  have 

am  -r-  a"  =  1,  when  m—n. 

E.g.,  a?  -r-  a2  =  1. 

EXERCISES  XVI. 

Express  each  of  the  following  quotients  as  a  single  power : 
1.   25  +  2.       2.   35^-32.      3.   x?  +  x2.  4.   a6 --a4. 

5.   x7-s-x\      6.    a6-=- a5.      7.   (—a)6  -7-  a5.     8.   (3  x)5  -5- (3  x). 
9.    (ab)7+(-aby.        10.   5n--53.  11.   an+1  -f-  a. 

1  o       n>n+7    .     n,n  -i  o       />»a+3    .     /yia+1  -i  4.       /v2n    .    /-»w— 1 

J.^.      X         -r"  X  .  AO.     X          -r-  «<C        .  i*.      tt'      T~  C*        . 


Division  of  Monomials  by  Monomials. 
38.   Ex.l.   12a-r-4  =  (12--4)  xa  =3xa  =  3 
Ex.  2.    _27a;7-f-3^  =  (-27--3)x(a;7--^)  = 
Ex.  3.   15  a*b2  -K-  5  a&2)  =  [15  -h  (-  5)]  x  (a3  --  a)  x  (62  -  62) 
=  -3  a2. 

Ex.  4.    -  16  ^w+1  ^-  (  -  8  x™yn-1) 

=  [-  16  -  (-  8)]  x  (x2m  -  aJ")  x  (yn+l  -J-  r'1) 


These  examples  illustrate  the  following  method  of  dividing 
a  monomial  by  a  monomial  : 

Multiply  the  quotient  of  the  numerical  coefficients  by  the  quotient 
of  the  literal  factors. 

EXERCISES  XVII. 

Divide 

1.  2.  3.  4.  5. 

2)6  a.     5)  -10  a.     4)  -16m.      —5y)-2oy.      —  7m)-49my. 


37-40] 


DIVISION. 


71 


6.  5x*  by  x. 

8.  25m5  by  -5m2. 

10.  6abc3  by  —Sac. 

12.  30  aty4  by  5x*if. 

14.  35a7610c13  by  -5aW. 

16.  15  (a  +  6)  by  3  (a  +  6). 

18.  10  a-"65  by  -  5  anb3. 

20.  #2n-ym+2  by  icw+yw-3. 

Simplify 

22.   a-V  -f-  (-  aa;3)  x  2  axy. 
24.   6  xm+lyn~l  -* 


7.  —6  a3  by  2  a. 

9.  -4a26  by  -2  a. 

11.  —  9a36  by  3a2b. 

13.  --15a567  by  -3a365. 

15.  12mVp8  by  -2m*»y. 

17.  25z2(a;+l)3  by  -5a 

19.  •-  27  a;B+ym  by  -  9  xy2m. 

21.  a*-1^-2  by  an-3bn~*. 


23. 


Division  of  a  Multinomial  by  a  Monomial. 

39.  If  the  indicated  operation  within  the  parentheses  in  the 
quotient  (8  +  6  —  4)  -5-  2  be  first  performed,  we  have 

(8  +  6  -  4)  -«-  2  =  10  -f-  2  =  5. 

But  if  each  term  within  the  parentheses  be  first  divided  by 
2,  and  the  resulting  quotients  be  then  added,  we  have 

8^-2  +  6^-2-4-j-2  =  4  +  3-2  =  5,  as  above. 

Therefore   (8  +  6-4)-i-2  =  8-5-2  +  6-5-2-4-5-2. 

This  example  illustrates  the  following  method  of  dividing  a 
multinomial  by  a  monomial  : 

Divide  each  term  of  the  multinomial  by  the  monomial,  and  add 
algebraically  the  resulting  quotients. 

That  is, 

(6+c  —  d)-t-a  =  b-t-a  +  c-:-a  —  d  -t-a. 

This  principle  is  called  the  Distributive  Law  for  division. 

40.  Ex.  1.   Divide  6  a,-2  -12  a  by  3x. 
We  have 


Ex.  2.   (4  a2"1'1  -  8  a3m+1)  -s-  4  am~l 

=  4  a2m-l  ^_  4  am-l  _  §  a3m+l  _j_  4  a«-l  =  am  _ 


72  ALGEBRA.  [Cn.  Ill 

EXERCISES  XVIII. 

Divide 

1.  5  +  10  a  by  5.  2.   4  a  +  8  b  by  -  4. 

3.  ax  -f  &a;  by  a?.  4.   3  a2  —  6  ab  by  —  3  a. 

5.  21  a26  -  14  a&2  by  -  7  ah. 

€.  8  am2  -  2  crm  +  4  aW  by  2  am. 

7.  25  (a  +  6)3  -  20  (a  +  6)  by  5  (a  +  6). 

8.  2  (a?  -  ?/)3  -  12  a  (a  -  y)4  -  6  (a?  -  2/)6  by  2  (aj  -  y)z. 

Simplify 

9.  2  a2  -(a3  -3  a)  --a. 

10.  (6x-4x2)-f-2x-  (-2x*y  +  3xy)-t-xy. 

11.  (a6  -  a26  +  3  o86)  -*•  ab  -  (4  a3  -  4  a2)  -s-  2  a. 

Divide  9  a*a?  -  6  aV  +  12  aV  by 

12.  3  a2.        13.    -3.T3.  14.    ax2.  15.    -  f  aV. 

Divide  105  aW  -  21  dW  +  42  a'dV  by 

16.    7  a3.        17.    -3a3^2.        18.    -  a26c3.  19.    faW. 

Divide  15  a^n+y  -  12  x2n+y  -  18  a^+Y  by 

20.   3«w.        21.    -5xn+l.     22.    -3^n+-       23. 


Zero  in  Division. 

41.   Since   0  -f-  JV  =  (a  —  a)  -*•  JV,  by  definition  of  0, 

=  a-^-a--^=0. 
We  have      0  -h  /If  =  0,  when  ^  is  not  equal  to  0. 

Observe  that  this  relation  is  proved  only  when  N  is  not 
equal  to  0. 


40-44]  DIVISION.  73 


Division  of  a  Multinomial  by  a  Multinomial. 

42.  The  division  of  one  multinomial  by  another  is  performed 
in  a  way  similar  to  that  of  dividing  one  number  by  another  in 
Arithmetic. 

Ex.   Divide  125  by  5. 

We  have 

125 | 5 


20  x  5  =  100  20  +  5  =  25 

125  -  20  x  5  =  ~25 
25 

The  work  is  equivalent  to  the  following  : 
125  -H  5  =  20  +  (125  -  20  x  5)  -=-  5  =  20  +  25  ->-  5  =  25. 

43.  The  number  20,  obtained  by  the  first  step  of  the  division, 
is  called  the  Partial  Quotient  at  that  stage.     It  is  the  greatest 
number  whose  product  by  the  divisor  is  equal  to  or  less  than 
the  dividend. 

In  general,  if  D  be  the  given  dividend,  d  the  given  divisor, 
and  q  the  partial  quotient,  the  principle  used  above,  stated 
symbolically,  is  : 

0^d=q  +  (D-qd)^d. 

44.  The  following  example   illustrates  the   application  of 
this  principle  in  dividing  one  multinomial  by  another. 


Ex.   Divide  x*  +  3x  +  2  by 

We  have 

(1) 
(2) 
=a;+(2a;+2)-Ka;+l)  (3) 


=a?+2+0-s-(a;+l) 

=x+2,  since  0-s-(a;+l)=0. 


74  ALGEBRA.  [Cn.  Ill 

We  take  the  quotient  of  the  term  containing  the  highest 
power  of  x  in  the  dividend  by  the  term  containing  the  highest 
power  of  x  in  the  divisor  as  the  partial  quotient  at  each  step. 

The  work  may  be  arranged  more  conveniently  thus : 

a?  +  l 

x  +  2,  quotient. 

to  be  subtracted  from  x2  +  3  x  +  2  ;   see  (1) 
and  (2)  above. 

Remainder  to  be  divided  by  x  +  1 ;  see  (3)  above. 
2  (x  +  1)  to  be  subtracted  from  2  x  +  2  ;  see  (4). 

45.  The  method  of  applying  the  principle  of  Art.  43  to  the 
division  of  multinomials,  as  illustrated  by  this  example,  may 
be  stated  as  follows : 

Arrange  the  dividend  and  divisor  to  ascending  or  descending 
powers  of  some  common  letter,  the  letter  of  arrangement. 

Divide  the  first  term  of  the  dividend  by  the  first  term  of  the 
divisor,  and  write  the  result  as  the  first  term  of  the  quotient. 

Multiply  the  divisor  by  this  first  term  of  the  quotient,  and  sub- 
tract the  resulting  product  from  the  dividend. 

Divide  the  first  term  of  the  remainder  by  the  first  term  of  the 
divisor,  and  write  the  result  as  the  second  term  of  the  quotient. 

Multiply  the  divisor  by  this  second  term  of  the  quotient,  and 
subtract  the  product  from  the  remainder  previously  obtained. 
Proceed  with  the  second  remainder  and  all  subsequent  remainders, 
in  like  manner,  until  a  remainder  zero  is  obtained,  or  until  the 
highest  power  of  the  letter  of  arrangement  in  the  remainder  is  less 
than  the  highest  power  of  that  letter  in  the  divisor. 

In  the  first  case  the  division  is  exact ;  in  the  second  case  the 
quotient  at  this  stage  of  the  work  is  called  the  quotient  of  the 
division,  and  the  remainder  the  remainder  of  the  division. 

46.  Ex.  1.   Divide  x2  -  4  x  -  5  by  x  -  5.     We  have 

x2  —  4  x  —  5  !B— 5 
x2  -  5  x 

x-5 

x-5 


44-46]  DIVISION. 

Ex.  2.   Divide 

a*b  -  15 64  -f  19  ab3  +  a4  -  8  a2b2  by  a2  -  5  62  +  Sab. 
Arranging  to  descending  powers  of  a,  we  have 


75 


^  4.    a36  -  8  aL7>2  +  19  aW  -  15  64 
_  5  a.26a 


a2  +  3  afr  -  5  62 


-  2  a36  -  3  a26*  +  19  a&3 


a2  -  2  a&  +  3  62 


3a262+ 
3a262+ 


—  156* 


Ex.  3.   Divide  8  x3  -  y3  by  2  ^  +  4  a2  +  ^/2. 

Arranging  the  divisor  to  descending  powers  of  x,  we  have  : 


+  2  a;? 


—  4  a?2?/  —  2  xy'2  —  y3 


2x  -      y 


Observe  that  the  remainder  after  the  first  partial  division  is 
arranged  to  descending  powers  of  x. 

Ex.4.    Divide   12a"+1  +  San-  45  a-1  +  25aw~2  by  6a-5. 
We  have 

6a-5 

12 an+1  -  10 an 

18  a*— 46  a—1 

18  an  -  15  a"-1 


—  30  an~l  -f  25  an~2 
_  3Q  fl«-i  +  25  a""2 

Ex.  5.     Divide       a,-3  +  (a  +  6  +  c)  cc2  4-  (a&  +  ac  +  6c)  CP  +  abc 
by  a?2  +  (a  +  6)  x  +  06. 

We  have 

x3  +  (a  -f-  b  +  c)  x2  +  (ab  +  ac  +  be)  x  +  a&c  I  x2  +  (a  - 
or  +  (a  +  6       )  x2  +  a&a  I  a?  4-  c 


ex2  +  (ac  +  be)  x  +  a&c 
ex2  H-  (ac  +  6c)  x  -f-  # &c 


76  ALGEBRA.  [Cn.  Ill 

EXERCISES  XIX. 

Find  the  values  of  the  following  indicated  divisions : 

1.    (a-3  +  2  a  +  1) -*- («  +  1).        2.   (x2+  llaj+30)-5-(aj+5). 
3.    (a-a  +  x  _  90)  _=_(£_  9).         4.  .(a-2  _  5  x  +  6)  -s-  (x  -  3). 
5.    (a2  +  7.T-44)-f-(x  +  ll).       6.   (a*-3aj-40)-s-(aj-8). 
7.    (3^-13^-10) -=-(3x4-2).     8.   (2  a2  +  a-6)-f-(2  a-3). 
9.    (15x2-7a-2)-:-(5a+l).    10.   (6  z2- 23  a +20)-=- (2  #-5). 

11.  (oj»- 4  as2 -20  a +  3) -s- (a: +  3). 

12.  (a;8  -  7  x2  +  13  x  -  15)  -«-  (x  -  5). 

13.  (4  x3  -  3  a;2  -  24  a;  -  9)  -s-  (a?  -  3). 

14.  (3a?-13ajs  +  23aj-21)-fr-(3o;-7). 

15.  (18  a?  +  7  x  +  10)  -s-  (3  x  +  2). 

16.  (50  x3  -  23  x  -j-  6)  -s-  (5  x  -  2). 

17.  (a2  +  2  aZ>  +  62)  -*-  (a  +  6). 

18.  2x2  +  6a2  +  7aa?_=-2a;  +  3a.. 
19. 

20. 

21.    (8  x2/  -  65  z?/z2  -  63  z4)  -5- 

22. 

23. 

24. 

25.  (ac-  6  a4  +  9  a2  -  4) -s-  (a2-!). 

26.  (21  a66  +  20  64  -  22  tfb3  -  29  a462)  -  (3  a26  -  5  62). 

27.  (a?  +  8  x2  +  9  x  -  18)  -s-  (x2  +  5  x  -  6). 

28.  (a;4  +  ar5-4x2  +  5a-3)--(a,'2  +  2.T-3). 

29.  (6  .T4  -  x5  -  11  x2  -  10  x  -  2)  -«-  (2  a2  -  3  x  -  1). 

30.    (a?  -  1) -*- (a2  +  «  +  1).         31.    (a3  +  8)  -*-  (a2  -  2a  +  4). 
32.    (^6-642/3)-(x2-4y).       33.    (aV  +  /)  -5-  (ax  +  y). 


46]  DIVISION.  77 

34.  (04  _|_  x2  +  1)  -  (x2  -  X  +  1). 

35.  (aV  4-  64  a?)  -r-  (4  ax  +  aV  +  8). 

36.  (4  a4  _  25  c4  -  30  6-c-  -  9  b4)  -  (2  a2  +  5  c2  +  3  62). 

37.  (27x4-6cV44c4)-f-(c2-6cx4-9z2). 

38.  (8  aV  +  32  a6  +  1  ?i6)  -5-  (4  an  +  ?i2  +  4  a2). 

39.  (16  a4Z>2+9  a*b*-l2  as6s-8  a56+3  a6) -(a4+3 a262-2  a36). 

40.  (28  a5c  -  26  a3c3-  13  a4c24-  15  a2c4)  -s-  (2  a2c2+  7  a8c  -  5  ac3). 

41.  (81  z*  -  90  64^4  +  81  66^2  -  20  Z>8)  --  (9  24  +  9  b2z2  -  5  64). 
42. 

43. 

44.  (a2  +  2  a&  +  b2  -  x*  +  4a?y  -  4  ?/2)-f-(a  +  6  -  «  +  2  y). 

45.  (a2  +  2  ac  -  62  -  2  6d  +  c2  -  d2)-s-(a  +  c  -  b  -  d). 

Find  the  values  of  the  following  indicated  divisions : 

46.  [a?2  +  (a  +  1)  #  +  a]  -*-  (*  +  a)- 

47.  [aj2-(a  +  6)aj  +  a*]-s-(a;-6). 

48.  [cor  —  (abc  +  1)  x  +  a6]  -?-  (x  —  aft). 

49.  [(6  +  c) x-  —  box  +  aj3  —  6c  (6  4-  c)]  -=- («2  —  6c). 

50.  [or3  +(a  +  b  +  c)^2  +(a&  +  ac  +  &c)#  4-  a6c]-i- (aj  +  6). 

51.  [a?  —  (c/  4-  b  4-  c) x2  4-  (a&  4-  ac  4-  6c)  #  —  a&c]  -s-  (a?  —  c). 

52.  (6  &  -  25  x2"  4-  27  aJ"  -  5)  -  (2  xw  -  5). 

53.  (6  ^  -  11  x4w  4-  23  ar3w  4- 13  x2w  -  3  i»w  +  2)  -(3  xw  4-  2). 

54.  (6  x*n+l  -  29  tfn  4-  43  x2"-1  -  20  x2M~2)  -%-  (2  xn  -  5  x'1-1). 

55.  (1  4-  afa  -  2  a3*)  -=-  (3  a2a5  +  2  a31  +  2  a*  4-  a4*  4- 1). 
56. 

57. 

58.  (-  T9e  ^6  4-  tt2^4  -  f  a4^2  +  J  a«)-(f  ^  -  |  a2x  + 

59-  (t  «4  +  tV  &4  +  tfV  -  if  c4)-(|  a2  4-  i  &2  -  |  c2). 


78  ALGEBRA.  [Cn.  Ill 

47.  In  the  equation     D  +  d  =  q  +  (D  —  qd)  +  d, 

I)  —  qd  is  the  remainder  at  any  stage  of  the  work,  and  q  is  the 
corresponding  partial  quotient.  If,  for  brevity,  we  let  R  stand 
for  the  remainder  at  any  stage,  we  have 

D  +  d=q  +  R  +  d.  (1) 

That  is,  the  result  of  dividing  one  number  by  another  is  equal 
to  the  partial  quotient  at  any  stage,  plus  the  remainder  at  this 
stage  divided  by  the  given  divisor. 

E.g.,  29  -=-6  =  4  +  5-=-6  =  4  +  £; 

(a*  _  x  +  2)  -  (x  +  1)  =  (aj  -  2)  +  4  -  (x  +  1). 

48.  If  both  members  of  the  equation 

D  -H  d  =  q  +  R  -r-  d 
be  multiplied  by  d,  we  have 
D-r-d  X  d  =  (q 

=  qd  +  R  -j-  d  x  d 
=  qd  4-  R,  since  •*•  d  x  d  =  -T-  1. 
Therefore,  D  =  qd  +  R. 

That  is,  the  dividend  is  equal  to  the  product  of  the  quotient  at 
any  stage  and  the  divisor,  plus  the  remainder  at  this  stage. 

E.g.,  29  =  4  x  6  +  5,  and  tf  -  x  +  2  =  (x  -  2)  (x  +  1)  -f  4. 

EXERCISES  XX. 

Find  the  remainder  of  each  of  the  following  indicated  divi- 
sions, and  verify  the  work  by  applying  the  principle  of  Art.  48  : 


1.    (or2  -  7  a  +  11)  -:-  (a  -  2).       2.    (3aj8  +  5*-9)-5-(oj-4). 

3.    (a8  -17  x2  +  15x-13)H-(2o;-5). 

4. 


CHAPTER   IV. 

INTEGRAL   ALGEBRAIC  EQUATIONS. 

We  will  now  distinguish  between  two  kinds  of  equations. 

Identical  Equations. 

1.  An  example  of  the  one  kind  is  : 

(a  +  6)  (a  -  6)  =  a2  -  62. 

The  first  member  is  reduced  to  the  second  member  by  per- 
forming the  indicated  multiplication. 

2.  Such  an  equation  is  called  an  Identical  Equation,  or  more 
simply,  an  Identity. 

3.  Notice  that  identical  equations  are  true  for  all  values 
that  may  be  substituted  for  the  literal  numbers  involved. 

E.g.,  if  a  =  5  and  6  =  3,  the  above  equation  becomes 
8  x  2  =  25  -  9,  or  16  =  16. 

Conditional  Equations. 

4.  An  example  of  the  second  kind  is : 

x  +  1  =  3. 

The  first  member  reduces  to  the  second  member,  when  x  =  2. 
It  seems  evident,  and  it  is  proved  in  School  Algebra,  Ch.  IV., 
that  x  -|- 1  reduces  to  3  only  when  x  =  2. 

5.  Such   equations    impose   conditions  upon    the    values  of 
the  literal  numbers  involved.      Thus,  the  equation  in  Art.  4 
imposes  the  condition  that  if  1  be  added  to  the  value  of  x, 
the  sum  will  be  3. 

79 


80  ALGEBRA.  [Cii.  IV 

A  Conditional  Equation  is  an  equation  one  of  whose  members 
can  be  reduced  to  the  other  only  for  certain  definite  values  of 
one  or  more  letters  contained  in  it. 

Whenever  the  word  equation  is  used  in  subsequent  work  we 
shall  understand  by  it  a  conditional  equation,  unless  the  con- 
trary is  expressly  stated. 

6.  An  Integral  Algebraic  Equation  is  an  equation  whose  mem- 
bers are  integral  algebraic  expressions  in  an  unknown  number 
or  unknown  numbers. 

E.g.,  3x?  —  4  =  2x,  and  f  #  +  5y  —  |  are  integral  equations. 

7.  The  Degree  of  an  integral  equation  is  the  degree  of  its 
term  of  highest  degree  in  the  unknown  number  or  numbers. 

8.  A  Linear  or  Simple  Equation  is  an  equation  of  the  first 
degree. 

E.g.,  x  -f-  1  =  6  is  a  linear  equation  in  one  unknown  number. 

9.  A  Solution  of  an  equation  is  a  value  of  the  unknown  num- 
ber, or  a  set  of  values  of  the  unknown  numbers,  which,  if 
substituted  in  the  equation,  converts  it  into  an  identity. 

E.g.,  2  is  a  solution  of  the  equation  x  +  1  =  3, 
since,  when  substituted  for  x  in  the  equation,  it  converts  the 
equation  into  the  identity  2  +  1=3. 

The  set  of  values  1  and  2,  of  x  and  y,  respectively,  is  a  solu- 
tion of  the  equation  x  -f-  y  =  3,  since  1  +  2  =  3  is  an  identity. 

10.  To  Solve  an  equation  is  to  find  its  solution. 

An  equation  is  said  to  be  satisfied  by  its  solution,  or  the  solu- 
tion is  said  to  satisfy  the  equation,  since  it  converts  the  equation 
into  an  identity. 

11.  When  the  equation  contains  only  one  unknown  number, 
a  solution  is  frequently  called  a  Root  of  the  equation. 

E.g.,  2  is  a  root  of  the  equation  x  +  1  =  3. 


5-13]  INTEGRAL   ALGEBRAIC   EQUATIONS.  81 

Equivalent  Equations. 

12.  Consider  the  solution  of  the  equation 

fa?-5  =  l.  (1) 

Adding  5  to  both  members, 

}*  =  6.  (2) 

Dividing  by  3,  %x  =  2.  (3) 

Multiplying  by  4,  x  =  8.  (4). 

It  is  evident  that  8  is  a  root  of  equations  (1),  (2),  (3),  and  (4). 

In  thus  applying  the  principles  of  Ch.  L,  Art.  17,  we  re- 
place the  given  equation  by  a  simpler  one,  which  has  the  same 
root,  this  equation  by  a  still  simpler  one,  which  again  has  the 
same  root,  and  so  on. 

Such  equations  as  (1),  (2),  (3),  and  (4)  are  called  Equivalent 
Equations. 

In  general,  two  equations  are  equivalent  when  every  solution 
of  the  first  is  a  solution  of  the  second,  and  every  solution  of  the 
second  is  a  solution  of  the  first. 

13.  It  is  important  to  notice  that  the  use  of  the  principles 
given  in  Ch.  L,  Art.  17,  may  lead  to  incorrect  results. 

Thus,  by  (iii.),  we  should  be  permitted  to  multiply  both 
members  of  an  equation  by  an  expression  which  contains  the 
unknown  number. 

E.g.,  the  equation  x  —  3  =  0  has  the  root  3. 

Multiplying  both  members  by  x  —  2,  we  obtain 
0-3)0-2)  =  0. 

This  equation  has  the  root  3, 
since  (3  -  3)  (3  -  2)  =  0  •  1  =  0 ; 

and  also  the  root  2, 
since  (2 -3)(2-2)  =  -1  -0  =  0. 


82  ALGEBRA.  [Cn.  IV 

But  2  is  not  a  root  of  the  given  equation,  since  2  —  3  does 
not  equal  0. 

That  is,  in  multiplying  both  members  by  x  —  2,  we  gained  a 
root  2.  Observe  that  this  root  is  the  root  of  x  —  2  —  0. 

The  derived  equation  is  therefore  not  equivalent  to  the  given 
one. 

Again,  by  (iii.),  we  should  be  permitted  to  multiply  both 
members  of  an  equation  by  0. 

Multiplying  both  members  of  x  —  3  =  0,  by  0,  we  have 

0(a;-3)  =  0. 

Any  number  is  a  root  of  this  equation,  since 
0(1  - 3)  =0,  0(2  -  3)  =  0,  0(3-3)  =  0,  0(4- 3)  =0,  etc. 

Finally,  by  (iv.),  we  should  be  permitted  to  divide  both 
members  of  any  equation  by  an  expression  which  contains  the 
unknown  number. 

E.g.,  the  equation  (x  —  1)  (x  -f- 1)  =  3  (a;  —  1), 
has  the  root  1,  since 

(1  - 1)  (1  + 1)  =  3  (1  - 1),  or  0  x  1  =  3  x  0,  or  0  =  0  j 
and  the  root  2,  since 

(2 - 1) (2  +  1)  =  3(2 -  1),  or  1x3  =  3x1. 
Dividing  both  members  by  x  —  1,  we  obtain 
x  +  1  =  3. 

This  equation  has  the  root  2  only,  and  not  the  root  1  of  the 
given  equation. 

That  is,  in  dividing  both  members  by  x  —  1,  we  lost  the  root  1. 
Observe  that  this  root  is  a  root  of  x  —  1  =  0. 

The  derived  equation  is  therefore  not  equivalent  to  the  given 
one. 

14.  The  correct  statements  of  the  principles  which  are 
applied  in  solving  equations  are,  therefore,  as  follows: 

(i.)  Addition  and  Subtraction.  —  TJie  equation  obtained  by 
adding  to,  or  subtracting  from,  both  members  of  an  equation  the 
same  number  or  expression  is  equivalent  to  the  given  one. 


13-14]          INTEGRAL   ALGEBRAIC   EQUATIONS.  83 

(ii.)  Multiplication  and  Division.  —  The  equation  obtained  by 
multiplying  or  dividing  both  members  of  an  equation  by  the  same 
number,  not  0,  or  by  an  expression  which  does  not  contain  the 
unknown  number  or  numbers,  is  equivalent  to  the  given  one. 

These  principles  are  proved  in  School  Algebra,  Ch.  IV. 

In  the  solutions  of  equations  in  the  preceding  chapters,  we 
multiplied  or  divided  only  by  Arabic  numerals.  Nevertheless, 
we  required  each  result  to  be  checked. 

EXERCISES   I. 

Solve  each  of  the  following  equations  : 

1.   x(x  +  3)  =  x(x-5).        2.   3x(x-5)  =  3x(x  +  2). 

3.  2(x  +  1)  -  3(x  +  1)  +  9(oj  +  1)  +  18  =  7(x  +  1). 

4.  5  (x  -  7)  -  4  (oj  -  7)  +  11  (oj  -  7)  =  10  +  2  (x  -  7). 

5.  _8(3aj-5)  +  5(3aj-5)-17-2(3a;-5)  =  3. 

6.  «(»  +  l)  +  a>(a  +  2)  =  (aj  +  3)(2a;-l). 

7.  (5  a -2)  (3  a -4)  =  (3  a +  5)  (5  a -6). 

8.  2z  +  2(    +  3  =  2(    +  2a;-5. 
9. 

10.  (16  a  +  5)  (9  # -f- 31)  =  (4  o,'  + 14^36  a; +  10). 

11.  arJ_.Tl-x-23-^     =  ic  +  l. 
12. 

13. 

14.    —  4  — 4{4  — 4[4  — 4(4  — a?)]}  = 

15. 

16.  4{4[4(4aj-3)-3]-3j-3  =  l. 

17.  3[5{5(a?-3)-3j-7]  =  2(aj  +  2 


84  ALGEBRA.  [Cn.  IV 

Problems. 

Pr.  1.  A  man  has  $  4.50  in  dimes  and  dollars,  and  he  has 
five  times  as  many  dimes  as  dollars.  How  many  coins  of  each 
kind  has  he  ? 

Let  x  stand  for  the  number  of  dollars. 

Then  5  x  stands  for  the  number  of  dimes. 

We  must  tirst  express  the  dimes  as  fractional  parts  of 
dollars,  or  the  dollars  as  multiples  of  dimes.  The  latter 
method  is  the  simpler.  Since  one  dollar  is  10  dimes,  x  dol- 
lars are  10  #  dimes. 

The  man  evidently  has  45  dimes. 

The  problem  states, . 

in  verbal  language :    ten  times  the  number  of  dollars  plus  the 
number  of  dimes  is  equal  to  45 ; 

in  algebraic  language :     10  x  -f  5  x  =  45, 

15^  =  45; 

whence  x  =  3, 

the  number  of  dollars. 

Then  5  x,  =  15,  the  number  of  dimes. 

Evidently  the  value  of  the  coins  is  3  +  i-J  dollars,  or  $  4.50. 

As  in  this  problem,  the  magnitudes  of  all  concrete  quantities 
of  the  same  kind  must  be  referred  to  the  same  unit ;  if  x  stand 
for  a  certain  number  of  yards,  then  all  other  distances  must 
likewise  stand  for  numbers  of  yards,  not  of  miles  or  of  feet. 

Pr.  2.  I  have  in  mind  a  number  of  six  digits,  the  last  one 
on  the  left  being  1.  If  I  bring  this  digit  to  the  first  place  on 
the  right,  I  shall  obtain  a  number  which  is  three  times  the 
number  I  have  in  mind.  What  is  the  number  ? 

Let  x  stand  for  the  number  which  is  composed  of  the  five 
digits  on  the  right  of  1. 

Then  the  original  number  is  100,000  +  x. 

When  1  is  moved  to  the  first  place  on  the  right,  each  digit 
in  x  is  moved  one  place  to  the  left.  Therefore,  the  resulting 
number  is  10  x  -f  1. 


14]  INTEGRAL   ALGEBRAIC   EQUATIONS.  85 

The  problem  states, 
in  verbal  language :  the  resulting  number  is  equal  to  three  times 

the  original  number ; 

in  algebraic  language  :     10  x  +  1  =  3  (100,000  +  x), 
whence  7  x  =  299,999, 

and  x  =  42,857. 

Therefore  the  required  number  is  142,857. 

Pr.  3.  A  man  asked  another  what  time  it  was,  and  received 
the  answer :  "  It  is  between  5  and  6  o'clock,  and  the  minute- 
hand  is  directly  over  the  hour-hand."  What  time  was  it? 

At  5  o'clock,  the  minute-hand  points  to  12  and  the  hour- 
hand  to  5.  The  hour-hand  is  therefore  25  minute-divisions  in 
advance  of  the  minute-hand. 

Let  x  stand  for  the  number  of  minute-divisions  passed  over 
by  the  minute-hand  from  5  o'clock  until  it  is  directly  over  the 
hour-hand  between  5  and  6  o'clock. 

Since  the  minute-hand  must  pass  over  25  more  minute- 
divisions  than  the  hour-hand  in  order  to  overtake  the  latter, 
the  number  of  minute-divisions  passed  over  by  the  hour-hand 
is  x  —  25. 

The  problem  states,  or  implies, 

in  verbal  language  :  the  number  of  minute-divisions  passed  over 
by  the  minute-hand  is  12  times  the  number  of  minute-divi- 
sions passed  over  by  the  hour-hand; 

in  algebraic  language  :  x  =  12  (x  —  25). 

From  this  equation  we  obtain  x  =  27^.  Consequently,  the 
two  hands  coincide  at  27T8T  minutes  past  5  o'clock. 

EXERCISES   II. 

1.  The    sum   of    three   consecutive    numbers   exceeds   the 
second  by  42.     What  are  the  numbers  ? 

2.  A  and  B  divide  a  sum  of  money.     A  receives  $  3  as  often 
as  B  receives  $5.     If  A  receives  $  3  x,  how  many  dollars  does 
B  receive  ? 


86  ALGEBRA.  [Cn.  IV 

3.  A  and  B  divide  $  1200.     A  receives  $  3  as  often  as  B 
receives  $  5.     How  many  dollars  does  each  receive  ? 

4.  The  length  of  a  room  is  four  times  its  width.     If  it  were 
12  feet  shorter  and  12  feet  wider,  it  would  be  square.     What 
are  the  dimensions  of  the  room  ? 

5.  A  man  travels  144  miles  by  train,  boat,  and  stage.     He 
travels  20  miles  farther  by  boat  than  by  stage,  and  three  times 
as  far  by  train  as  by  boat  and  stage  together.     How  many  miles 
does  he  travel  by  each  conveyance  ? 

6.  A  man  paid  a  debt  in  four  monthly  payments.     He  paid 
$45  more  each  month  than  the  preceding.     If  his  debt  was 
three  times  his  last  payment,  how  much  was  his  first  payment  ? 
How  much  was  his  debt  ? 

7.  In  a  number  of  two  digits,  the  tens'  digit  is  three  times 
the  units'  digit.     The  number  itself  exceeds  four  times  the 
units'  digit  by  54.     What  is  the  number  ? 

8.  In  a  number  of  two  digits,  the  tens'  digit  is  twice  the 
units'  digit.     If  the  digits  are  interchanged,  twice  the  resulting 
number   exceeds   the   original   number   by   9.      What   is   the 
number  ? 

9.  Three  boys,  A,  B,  and  C,  have  a  number  of  marbles. 
A  and  B  have  55,  B  and  C  have  62,  and  A  and  C  have  57. 
How  many  marbles  has  each  boy  ? 

10.  A  man,  wishing  to  give  alms  to  several  beggars,  lacks 
15  cents  of  enough  to  give  22  cents  to  each  one.     If  he  were 
to  give  20  cents  to  each  one,  he  would  have  1  cent  left  over. 
How  many  beggars  are  there  ? 

11.  A,  travelling  25  miles  a  day,  has  3  days'  start  of  B,  who 
travels  30  miles  a  day  in  the  same  direction.     After  how  many 
days  will  B  overtake  A  ? 

12.  The  sum  of  two   numbers   is  47,  and  their  difference 
increased  by  7  is  equal  to  the  less.     What  are  the  numbers  ? 


15]  INTEGRAL   ALGEBRAIC   EQUATIONS.  87 

13.  The  sum  of  three  consecutive  even  numbers  exceeds  the 
least  by  42.     What  are  the  numbers  ? 

14.  Atmospheric  air  is  a  mixture  of  four  parts  of  nitrogen 
with  one  of  oxygen.     How  many  cubic  feet  of  oxygen  are  there 
in  a  room  12  yards  long,  5  yards  wide,  and  17  feet  high  ? 

15.  A  merchant  paid  $  7.50  in  an  equal  number  of  dimes 
and  five-cent  pieces.     How  many  coins  of  each  kind  did  he 
pay? 

16.  A  man  has  $5.70  in  dimes  and  quarters,  and  he-has 
6  more  quarters  than  dimes.     How  many  coins  of  each  kind 
has  he  ? 

17.  In  my  right  pocket  I  have  as  many  dollars  as  I  have 
cents  in  my  left  pocket.     If  I  transfer  $6.93  from  my  right 
pocket  to  my  left,  I  shall  have   as  many  dollars  in  my  left 
pocket  as  I  shall  have  cents  in  my  right.     How  much  money 
have  I  in  my  left  pocket  ? 

18.  One  barrel  contained  36  gallons,  and  another  60  quarts, 
of  wine.     From  the  first  three  times  as  much  wine  was  drawn 
as  from  the  second;    the  first  then  contained  twice  as  much 
wine  as  the  second.     How  much  wine  was  drawn  from  each  ? 

19.  A  regiment  moves  from  A  to  B,  marching  18  miles  a 
day.     Two  days  later  a  second  regiment  leaves  B  for  A,  and 
marches  26  miles  a  day.     At  what  distance  from  A  do  the 
regiments  meet,  A  being  212  miles  from  B  ? 

20.  A  man  travels  3  miles  in  one  hour.     During  the  first 
half-hour,  he  goes  10  yards  farther  every  minute  than  during 
the  second  half-hour.     How  many  yards  a  minute  does  he  go 
the  first  half-hour  ? 

21.  The  greatest  of  three  vessels  holds  28  gallons  more  than 
the  second,  and  45  gallons  more  than  the  third.     If  the  con- 
tents of  the  second  and  third,  when  full,  are  poured  into  the 
first,  when  empty,  the  latter  will  lack  8  gallons  of  being  filled. 
What  is  the  capacity  of  each  vessel  ? 


88  ALGEBRA  [Cn.  IV 

22.  A  father  leaves  $  25,800  to  his  four  sons.     The  first 
receives  twice  as  much  as  the  second,  less  $  300 ;  the  second 
three  times  as  much  as  the  third,  less  $  600  ;  and  the  third 
four  times  as  much  as  the  fourth,  less  $  900.      How  many 
dollars  does  each  son  receive  ? 

23.  Two  bodies  move  from  the  same  point  in  the  same  direc- 
tion, one  at  the  rate  of  24  feet  a  minute,  the  other  at  the  rate 
•of  30  feet  a  minute.     If  the  second  starts  35  minutes  after  the 
iirst,  where  will  it  overtake  the  first  ?     When  will  the  distance 
between  them  be  270  feet  before  they  meet  ?     When  270  feet 
after  they  meet  ? 

24.  A  child  was  born  in  November.     On  the  10th  of  Decem- 
ber the  number  of  days  in  its  age  was  equal  to  the  number  of 
days  from  the  1st  of  November  to  the  day  of  its  birth,  inclu- 
sive.    What  was  the  date  of  its  birth  ? 

25.  A  person  attempts  to  arrange  a  number  of  coins  in  the 
form  of  a  square.     On  the  first  attempt,  lie  has  31  pieces  left 
over.     When  he  adds  2  to  each  side  of  his  square,  he  lacks  25 
coins  of  enough  to  complete  this  square.     How  many  coins 
has  he  ? 

26.  In  a  certain  family  each  son  has  as  many  brothers  as 
sisters,  but  each  daughter  has  twice  as  many  brothers  as  sisters. 
How  many  children  are  in  the  family  ? 

27.  A  merchant's  investment  yields  him  yearly  33 J%  profit. 
At  the  end  of  each  year,  after  deducting  $  1000  for  personal 
•expenses,  he  adds  the  balance  of  his  profits  to  his  invested 
•capital.     At  the  end  of  three  years  his  capital  is  twice  his 
original  investment.     How  much  did  he  invest  ? 

28.  I  have  in  mind  a  number  of  four  digits,  the  first  one  on 
the  right  being  2.     If  I  bring  this  digit  to  the  last  place  on  the 
left,  I  shall  obtain  a  number  which  is  less  than  the  number  I 
have  in  mind  by  2106.     What  is  the  number? 

29.  At  what  time  between  3  and  4  o'clock  will  the  minute- 
hand  of  a  watch  be  directly  over  the   hour-hand  ?     At  what 
time  between  9  and  10  o'clock  ? 


CHAPTER  V. 

TYPE-FORMS. 

1.  We  shall  in  this  chapter  consider  a  number  of  products 
and  quotients  which  are  of  frequent  occurrence.     They  enable 
us  to  shorten  work  by  writing  similar  products  and  quotients 
without  performing  the  actual  multiplications  and  divisions. 
They  are  called  Type-Forms. 

TYPE-FORMS  IN  MULTIPLICATION. 
The  Square  of  a  Binomial. 

2.  By  actual  multiplication,  we  have 

« 

(a  +  b)2=(a  +  b)(a  +  b)  =  a2  +  ab  +  6a  +  62  =  o2  +  2  a6 +  62. 

That  is,  the  square  of  the  sum  of  two  numbers  is  equal  to  the 
square  of  the  first  number,  plus  twice  the  product  of  the  two  num- 
bers, plus  the  square  of  the  second  number. 

E.g.,        (2x  +  5y)2  =  (2  xf  +  2 (2 oj)(5 y}  +  (5 yf 


3.   By  actual  multiplication,  we  have 
(a  -  6)2  =  (a  -  b)  (a  -  6)  =  a2  -  ab  -  ba  +  62  =  a2  -  2  ab  +  b\ 

That  is,  the  square  of  the  difference  of  two  numbers  is  equal  to 
the  square  of  the  first  number,  minus  twice  the  product  of  the 
two  numbers,  plus  the  square  of  the  second  number. 

E.g.,        (3x-7y)2=  (3 *)2 -  2 (3 x)(7  y)  +  (7 y)2 
=  9  x2  -  42  xy  +  49  y2. 


90  ALGEBRA.  [Cn.  V 

4.  Observe  that  this  type-form  is  equivalent  to  that  of  Art.  2, 
since  a  —  b  =  a  +  (—&). 

E.g.,       (3x-7y)2  =  (3x)2  +  2 (3 a?)  (-  1  y)  +  (-  7*,)2 

=  9  x2  —  42  x?/  -f  49  ?/2,  as  above. 

The  signs  of  all  the  terms  of  an  expression  which  is  to  be 
squared  may  be  changed  without  changing  the  result. 
For,  (a  -  b)2  =  [  -  (b  -  a)]2  =  (b  -  of 

5.  In  applying  the  type-forms  in  this  Chapter,  it  will  be 
necessary  to  raise  a  monomial  to  any  required  power. 

We  have 

(5  a3&4)2=  5  -  5  asasb4b*=  52a3+3&4+4=  52a2x362x4=  25  a668. 

That  is,  to  square  a  monomial : 

Square  the  numerical  coefficient,  and  multiply  the  exponent  of 
each  literal  factor  by  2. 

In  general,  to  raise  a  given  monomial  to  any  required  power : 
Raise  the  numerical  coefficient  to  the  required  power,  and  mul- 
tiply the  exponent  of  each  literal  factor  by  the  exponent  of  the 
required  poiver. 

E.g.,  (3  ab2)s  =  3Wx8  =  27  o868. 

EXERCISES   I. 

Write,  without  performing  the  actual  multiplications,  the 
values  of : 

1.    (a  +  1)2.  2.  (x-3)2.  3.  (a  +  5)2. 

4.    (a_4)2.  5.  (3x  +  2)2.  6.  (4-oz)2. 

7.    (mn  +  6)2.  8.  (ab-S)2.  9.  (xy  +  z)2. 

10.    (4z2-3)2.  11.  (3xy  +  5z)2.  12.  (2ab-6bc)*. 

13.    (xy'2-3x2y)2.       14.  (2a~by-  9c2)2.  15.  (4  a263  -  8  c4)2. 

16.    (zn  +  l)2.  17.  (xm-yn)2.  18.  (an+1  +  a'1-1)2. 

Simplify  the  following  expressions  : 

19.    a*  +  V-(a-V)*.  20.    (x-y)2-(x  +  y)2. 

21.  x2  +  y2  -  4:x  +  6y  +  3,  when  x  =  a -f  1,  y  =  a-2. 

22.  (a  +  b  -  c)(a  +  6)  +  (a  -  6  +  c)(a  +  c)  +  (6  +  c  -  a)  (6  +  c). 


4-6]  TYPE-FORMS   IN   MULTIPLICATION.  91 

Verify  the  following  identities  : 

23.  (a2  +  b2)  (or  +  y2)  -  (ax  +  by)2  =  (ay  -  bx)2. 

24.  a2  +  b2  +  4c2  +  2ab  +  86c  =  4  (a  +  c)2,  when  b  =  a. 

Product  of  the  Sum  and  Difference  of  Two  Numbers. 
6.   By  actual  multiplication,  we  have 

(a  +  6)  (a  -  b)  =  a2  -  ab  +  ba  -  b2  =  a2  -  62. 

That  is,  the  product  of  the  sum  of  two  numbers  and  the 
difference  of  the  same  numbers,  taken  in  the  same  order,  is  equal 
to  the  square  of  the  first,  minus  the  square  of  the  second. 

Ex.  1.    (2x  +  3y)(2x-3y)  =  (2x)2-(3yy  =  4:X2-$y2. 

The  product  of  two  multinomials  can  frequently  be  brought 
under  this  type-form  by  properly  grouping  terms. 

Ex.  2.    (x*  +  x  +  I)(x2-x  +  1)  =  [>2+  1)+  x- 


Ex.3.       x- 


EXERCISES   II. 

Write,  without  performing  the  actual  multiplications,  the 
values  of: 

1.    (0  +  2)(a-2>  2.  (a  -6)  (a  +  6). 

3.    (m  +  9)(m-9).  4.  (2  a  +  1)  (2  a  -  1). 

5.    (5  a  -7)  (5*  +  7).  6.  (9-6«)(9  +  5a;). 

7.    (2a  +  36)(2a-36).         8.  (5x  —  6y)(5x  +  6y). 

9.    (-8w+5n)(8m+5?i).    10.  (a&  +  1)  (a&  -  1). 

11.    (3  ax  —  4)  (3  ax  +  4).         12.  (—  ay  +  z)  (ajy  +  z). 

13.    (-2a&4-c)(2a6  +  c).     14.  (5  a?y  -  3  z)  (5  xy  +  3  2). 


92  ALGEBRA.  [Cn.  V 

15.    02  +  1)(>2-1).  16.    (3  a3  +  4)  (3  a3  -  4). 

17.    (5a4-26)(5a44-  26).     18. 

19.  (3  a"  4-  5)  (3  a"  -5). 

20.  (-5 #n+1  4-  9 a?"-1)  (5  aj"+1  +  9  a;'1-1). 

21.  [a2  4-  6  (a  +  6)]  [a2  -  6  (a  +  6)]. 

22.  (aj  +  y  +  5)(aj  +  y-5). 

23.  (4  a  -  3  b  -  7)  (4  a  -  3  b  +  7). 

24.  (^  +  /  +  z2)(-*2  +  2/2  +  Z2)' 

25.  (a2  -  ab  4-  62)  <X  4-  ab  +  62). 

26.  <>2  +  2a;-l)(>2-2z-l). 

27.  (x4  -  x2  4-  1)  (V  4-  a^  -  1). 

28.  (-c^- 

Simplify  the  following  expressions  : 

29.  (l 

30.  (2x 

31.  (a;  -  3)  (a -1)  (a? +  !)(«  + 3). 

32.  (a  —  x)(a  +  x)  (a2  +  x~)  (a4 

33.  (aj'- 

34.  (ajz- 

35.  (a  +  &  -  c)  (a  +  c  -  6)  (6  +  c  -  a)  (a  +  &  -f  c). 

The  Product  (x  +  a)  (x  +  6) . 
7.   By  actual  multiplication,  we  have 
(JT  +  a)(jf  +  6)  =  *2  +  a*  +  6jr  +  a6  =  jr2  +  (a  +  6)jr  +  a^ ; 
(JT  +  a)(jr  —  6)  =  jr2  +  ajr  —  6jr  —  ab  =  x1  +  (a  —  b)x  —  ab  ; 
(JT  —  a)(jr  —  6)  =  x2  —  ax  —  bx  +  ab  =  x2  —  (a  +  6)jf  -f  aA. 

We  thus  derive  the  following  method  for  multiplying  two 
binomials  which  have  a  common  first  term : 

The  first  term  of  the  product  is  the  square  of  the  common  first 
terms  of  the  binomials. 


6-7]  TYPE-FORMS   IN  MULTIPLICATION  93 

The  coefficient  of  the  second  term  of  the  product  is  the  algebraic 
sum  of  the  second  terms  of  the  binomials. 

The  last  term  of  the  product  is  the  product  of  the  last  terms  of 
the  binomials. 

Ex.  1.   Write  the  product  (x  +  3)  (a;  +  7). 

The  first  term  is  x* ; 

The  second  term  is  (3  -f-  7)  a;,   =  10  x ; 

The  third  term  is  3  x  7  =  21. 

Therefore  (x  +  3)  (x  +  7)  =  x*  +  10  x  +  21. 

Ex.  2.   Write  the  product  (x  —  8)  (x  +  2). 

First  term :  x2 ;  second  term :  (—  8  +  2)  x,  =  —  6  a? ; 
third  term :   -  8  x  2  =  -  16. 

Therefore        (a;  -  8)  (x  +  2)  =  x2  -  6  x  -  16. 

Ex.  3.    Write  the  product  (a2  +  9)  (a2  -  3). 

First  term :  (a2)2,  =  a4 ;  second  term  :  (9  —  3)  a2,  =  6  a2 ; 
third  term:  9  x  (-3),  =-27. 

Therefore       (a2  +  9)  (a2  -  3)  =  a4  +  6  a2  -  27. 

Ex.  4.    Write  the  product  (x  —  5y)(x—7 y). 

First  term:  x2;  second  term :  (—5y  —  7y)x,  =  —  'L2xyJ 
third  term  :  —5yx(—7y),  =  35 y2. 

Therefore  (x  -  5  y)  (x  -  7  y)  =  x2  - 12  xy  -f  35  y2. 

EXERCISES   III. 

Write,  without  performing  the  actual  multiplications,  the 
values  of : 

1.   (aj  +  2)(aj  +  3).  2.  (x  +  2)  (x  -  3). 

3.    (x  -  2)  (x  -f  3).  4.  (aj-2)(aj-3). 

5.    (a?  +  5)  (a?  +  8).  6.  (a?  +  6)  (a?  -  8). 

7.    0-5)(«  +  8).  8.  0-5)0-8). 

9.    (8  +  m)(w  — 9).  10.  (5  +  a)  (a  -6). 

11.    (7 +  3  a?)  (7 -a?).  12.  (-  3  +  5  a)  (6  +  5  a). 

13.     a  +       aj  +  2.  14.  x  +  yx 


94  ALGEBRA.  [Cn.  V 

15.  (x-y)(x-2y).  16.  (ab  +  1)  (ab  -  3). 

17.  (#y  +  7)  (xy  -  8).  18.  (ab  +  3  c)  (a&  -  5  c). 

19.  (or'  +  8)  (or2  —  9).  20.  (afy  -  5)  (xhj  +  11). 

21.  (a#*  +  9  a)  O/  -  6  a).  22.  (a8  +  3  aft)  (a2  -  2  a&). 

23.  (an  +  2)  (an  -  5).  24.  (xTO+l-  3)(^w+1+  8). 

25.  (a  +  t  +  3)(a  +  &  —  7).  26.  (a  -  y  +  3z)  (x  -  y  -  5z). 

The  Product  (ax  +  6)(«r  +  </). 
8.    By  actual  multiplication,  we  obtain 

(ajr  +  6)  (CJT  +  (/)  =  acjr2  +  (at/  +  6c);r  +  bd. 
In  this  type-form  that  part  of  the  multiplication  which  gives 
the  middle  term  of  the  type-form  may  be  represented  concisely 
by  the  following  arrangement  : 

cx  +  d 


x 

ax  -f-  o 


(ad  -f-  bc)x 

The  products  of  the  terms  connected  by  the  cross  lines  are 
called  cross-products^  and  their  sum  is  the  middle  term  of  the 
given  trinomial. 

That  is,  the  product  of  two  binomials,  arranged  to  powers  of  a 
common  letter,  is  equal  to  the  product  of  the  first  terms,  plus  the 
sum  of  the  cross-products,  plus  the  product  of  the  last  terms. 

Ex.1.     7x 


EXERCISES   IV. 

Write,  without  performing  the  actual  multiplications,  the 
values  of: 


1.  (3a  +  l)(5a  +  2).  2.  (7x-  3)(3x-  1). 

3.  (5  a?  +  7)  (3  a?  -2).  4.  (2x  -  9)  (5  a?  +  1). 

5.  (2  a?  +  15)  (4  a?  -5).  6.  (11  a  -  3)  (9  a  +  7). 

7.  (2a  +  6)(3a-6).  8.  (2  a  -  b)  (3  a  +  b). 


7-9]  TYPE-FORMS  IN   DIVISION.  95 

9.    (3x-y)(2x-y).  10.  (la  +  3  b)  (5  a  +  26). 

11.    (6x-7y)(3x  +  2y).  12.  (5  a;  -  3z)(2x  +  5z). 

13.    (7y  +  2w)(82/-7w).  14.  (2ab-x)(3ab  +  x). 

15.    (5  mn  +  3p)  (6  mw  +  7jp).  16.  (9  m2  -  3)  (8-m2  +  11). 

17.  (3^  + 

18.  [3(a  + 

19.  2aJ- 


TYPE-FORMS   IN  DIVISION. 

Quotient  of  the  Sum  or  the  Difference  of  Like  Powers  of  two 
Numbers  by  the  Sum  or  the  Difference  of  the  Numbers. 

9.   By  actual  division,  we  obtain 

(a*-P)  +  (a  +  fy  =  a-6  and  (a2-  62)  -s-  (a  -A)  =  a  +  6. 

That  is,  £/ie  difference  of  the  squares  of  two  numbers  is  divisible 
by  the  sum,  and  also  by  the  difference  of  the  numbers.  The 
quotient  in  the  first  case  is  the  difference  of  the  numbers,  taken  in 
the  same  order,  and  in  the  second  case  is  the  sum  of  the  numbers. 

Ex.1.    (9- 
.  Ex.2. 


EXERCISES   V. 

Write,  without  performing  the  actual  divisions,  the  values  of  : 

1.   (aj«  _  1)  -«-  (a;  -  1).  2.  (25  -  x2)  -*-  (5  +  x). 

3.  (4a2-9)-(2a-3).          4.  ($  -  tff)  -  (i  +  xy). 

5.  (3-4  _  i)  +  (aJ  +  i).  6.  (4a4_62)^(2a2-6). 

7.  (16x2-9/)--(4^-32/).     8. 

9.  (4aV-2/8)^-(2a2^H-2/4).   10. 

11.  (cc2n  -  1)  -=-  (xn  -  1).  12.  (a4"  -  16  616)  -5-  (a2w  +  4 

13.  (a*H-»  _  4)  -s-  (aJ"+1  +  2).     14.  («8w  -  ft41*4 

15.  [(a  +  6)2-l]-(a  +  &  +  l). 

16.  [4  -  (a  +  6)2]  -«.  (2  -  a  -  6). 


96  ALGEBRA.  [Ce.  V 

17.  (a2  -  2  ab  +  b2  -  1)  -r  (a  -  b  +  1). 

18.  (a2  —  7i2  —  p2  4-  2  np)  -s-  (a  —  n  -f  _p). 

19.  (jp2  -  r2  -  4  -  4  r)  -5-  (p  -  r  -  2). 

20.  [(a2  +  2  a&  +  &2)  **  -  2/4]  -i-  [(a  +  6)  a?  +  2/2]- 

21.  (#4  +  2o;y  +  y*-z2  -  2zu  -  u2)  +  (x2  +  y2  +  u  +  z). 

22.  (a2  -  b2  +  2  62  -  2  a;s  +  a?  -  z2)  -s-  (a  -  a?  -  6  +  z). 

The  Sum  and  Difference  of  Two  Cubes 
10.   By  actual  division,  we  obtain 

(flS  +  fiS)  _,_  (fl  +  ^)  =  fl2  _  fl6  +  ^  (J) 

(a3  -  63)  -v-  (a  -  6)  -  a2  +  a6  +  62.  (2) 

That  is,  ^e  sitm  of  the  cubes  of  two  numbers  is  divisible  by  the 
sum  of  the  numbers.  The  quotient  is  the  square  of  the  first  num- 
ber, minus  the  product  of  the  numbers,  plus  the  square  of  the 
second  number. 

The  principle  contained  in  (2)  may  be  stated  in  a  similar 
way. 

Ex.1. 


Ex.  2.          (a12  -  b9)  -f-  (a4  -  ft3)  =  (a4)2  +  a46»  +  (63)2 

=  a8  +  a4^3  4-  &?• 

EXERCISES   VI. 

Write,  without  performing  the  actual  divisions,  the  values  of: 
1.    (l  +  a3)^_(i_|_a).  2.    (aj8-  8)  -!-(»-  2). 

3.    (w8  +  27)-f-(m  +  3).  4.    (64  —  a^)-r-(4  —  a?). 

5.    (216  +  o3)-i-(6  +  a).  6.    (8  a3  -27)  -s-(2  a  -3). 

7.    (ajy  +  l)-s-(«y  +  l).  8. 

9.    (125x3/-26) 

10.  (27a669-64c3)--(3a2^-4c). 

11.  (8  www8  -  p12)  (2  m5n  -  p4). 


9-11]  TYPE-FORMS  IN   DIVISION.  97 


12.    (a3"  +  l)-i-(an  +  l).  13. 

14.  (343  a^-3  +  /»)  -  (7  x™~1  +  7/2M). 

15.  [(*  +  7/)3-8]-(.T-f-t/-2). 

16.  [1+  (a-?/)3]  +  (!  +  £-?/). 

17.  [(a  -  b)6  -  8  c»]-i-[a2  +  62  -  2(a&  +  c)]. 

Sum  and  Difference  of  Like  Powers  of  Two  Numbers. 
11.   By  actual  division,  we  find  : 

(a*  _  64)-(a  +  6)=  a3  -  a2b  +  ab2  -  b3  ; 

(a*  _  64)-(a  -  6)-  a3  +  a2b  +  a^2  +  63  ; 

a4  +  64  is  not  divisible  by  either  a  +  b  or  a  —  b  ; 

(a*  +  65)  -(a  +  6)  =  a4  -  asb  +  a262  -  a63  +  64  ; 

a5  +  65  is  not  divisible  by  a  —  b. 

The  above  identities  and  those  of  Arts.  9-10,  illustrate  the 
following  principles  : 

(i.)  an  —  bn  is  divisible  by  a  —  b,  but  not  by  a  -{-b,  when  n  is 
odd. 

The  quotient  is 

a,-l  +  an-'2b  +  an-362  +  ...   +  ^n-3  +  afin-2  +  J—l^ 

(ii.)  aw  —  &n  is  divisible  by  both  a  +  b  and  a—b,  when  n  is  even. 
The  quotient,  when  a  -f-  b  is  the  divisor,  is 

an-l  _  an-2b  +  an-362  -----  ^n-3  +  ^n-2  _  &_1  . 

and,  when  a  —  b  is  the  divisor,  is 


(iii.)  an  +  6n  ^s  divisible  by  a  -f  6,  6w£  noi  by  a  —  b,  when  n  is 
odd. 

The  quotient  is 

a»-i  _  an-25  +  aw~362  -----  h  a2bn-3  -  abn~2  -f-  bn~l. 

(iv.)  an  +  6n  is  wo^  divisible  by  either  a  -\-b  or  a  —  b,  when  n  is 
even. 


98  ALGEBRA.  [Cn.  V 

12.  The  following  directions  will  be  helpful  in  writing  the 
quotients  of  these  type-forins  : 

(i.)  When  the  divisor  is  a  sum,  the  signs  of  the  terms  of  the 
quotient  alternate,  +  and  —  . 

(ii.)  When  the  divisor  is  a  difference,  the  signs  of  the  terms  of 
the  quotient  are  all  +. 

(iii.)  In  the  first  term  of  the  quotient  the  exponent  of  a  is  less 
by  1  than  its  exponent  in  the  dividend,  and  decreases  by  1  from 
term  to  term. 

(iv.)  The  exponent  of  b  is  1  in  the  second  term  of  the  quotient, 
and  increases  by  1  from  term  to  term. 

Observe  that  the  quotient  is  homogeneous  in  a  and  b,  of 
degree  less  by  1  than  the  degree  of  the  dividend. 

Ex.  1.    (x4  -  16y4)  +  (x  -2y) 


x(2  yY 


The  proofs  of  these  type-forms  are  given  in  School  Algebra, 
Ch.  V. 

EXERCISES  VII. 

Write,  without  performing  the  actual  divisions,  the  values  of  : 
1.    (a*  _  i)  _*.  (a-  +  i).  2.    (1  -  a4)  -=-  (1  -  a). 

3.    (m5  -  1)  -i-  (m  -  1).  4.    (32  +  n5)  -*-  (2  +  71). 

5.    (ae  __  &6)  _,_  (a  _  6^  6     (ar  +  &7)  _,_  (a  +  &). 

7.    (aw  _  &io)  _,_  (a  _  &).  a    (a11  +  b11)  -=-  (a  +  b). 

9.    (^  _  ^)  _*.  (a*  _  /).  10.    (xw  +  2/10)  -  (x2  +  y*). 

11.    (ay-bl*)  +  (a2y  +  b3).         12.    (a10/  -  z15)  -h  (on/  -  z3). 

13.  (xc- 

14.  (243 

15.    (»4n  -  y4")  -^  (a;  -  y).  16.    (a^n  -  1)  --  («n  -  1). 

17.    (1  +  a5"1)  -=-  (1  +  a).  18.    (a1Vn+614w)-r-(aV+62m). 


CHAPTER  VI. 

FACTORS   AND   MULTIPLES   OF   INTEGRAL  ALGEBRAIC 
EXPRESSIONS. 

INTEGRAL  ALGEBRAIC  FACTORS. 

1.  A  product  of  two  or  more  factors  is,  by  the  definition  of 
division,  exactly  divisible  by  any  one  of  them. 

An  Integral  Algebraic  Factor  of  an  expression  is  an  integral 
expression  which  exactly  divides  the  given  one. 

E.g.,  integral  factors  of  6  a?x  are  6,  a?x,  3  x,  2  a2,  etc. ; 
integral  factors  of  a2  —  62  are  a  +  b  and  a  —  b. 

2.  A  Prime  Factor  is  one  which  is  exactly  divisible  only  by 
itself  and  unity. 

E.g.,  the  prime  factors  of  6  a?x  are  2,  3,  a,  a,  x. 

A  Composite  Factor  is  one  which  is  not  prime,  i.e.,  which  is 
itself  the  product  of  two  or  more  prime  factors. 

E.g.,  composite  factors  of  6a?x  are  6,  ax,  2 a,  Sax,  etc. 

3.  Any  monomial  can  be  resolved  into  its  prime  factors  by 
inspection. 

E.g.,  the  prime  factors  of  4  a3b2  are  2,  2,  a,  a,  a,  b,  b. 

Multinomials  whose  Terms  have  a  Common  Factor. 

4.  From  Ch.  III.,  Art.  30,  we  have 

ab  +  ac  —  ad  =  a(b  +  (?  —  </).  (1) 

This  relation  may  be  called  the  Fundamental  Formula  for 
Factoring.     From  it  we  derive  the  following  method  for  find- 

99 


100  ALGEBRA.  [Cn.  VI 

ing  the  second  factor  of  a  multinomial  whose  terms  have  a 
common  factor  : 

Determine  by  inspection  the  remaining  factors  of  its  terms,  and 
take  their  algebraic  sum. 

5.  Ex.1.   Factor  2x?y-2xy2. 

The  factor  2xy  is  common  to  both  terms;  the  remaining 
factor  of  the  first  term  is  aj,  that  of  the  second  term  is  —  y, 
and  their  algebraic  sum  is  x  —  y. 

Consequently,  2  x2y  —  2  xy2  =  2xy(x  —  y). 

Ex.  2.   aW  +  abc  +  b~c  =  b  (ab  +  ac  +  be). 

6.  In  the  fundamental  formula  the  letters  a}  b,  c,  d  may 
stand  for  binomial  or  multinomial  expressions. 

Ex.  1.   Factor  a  (x  —  2  y)  +  6  (x  —  2  y). 

The  factor  x  —  2y  is  common  to  both  terms;  the  remaining 
factor  of  the  first  term  is  a,  that  of  the  second  term  is  b,  and 
their  algebraic  sum  is  a  +  b. 

Consequently  a  (x  —  2  y)  -f  b  (x  —  2  y)  =  (x  —  2  y)  (a  +  b). 
Ex.  2.   x*(l  -  m)  -  y\m  -  1)  =  ^(1  -  m)  +  y*(l  -  m) 


EXERCISES  I. 

Factor  the  following  expressions  : 

1.   5  x  +  o.  2.   ax  —  a.  3.   4  a3  —  6. 

4.    x*-2x\  5.   a2b  +  ab2.  6.    2  an  -In*. 

7.   Stfif-Sx-y3.      8.    12  asb3  +  3  a262.      9.    10  aV  -  15  a2iB*. 
10.   3a&  +  6ac-12ad.  11.   70  xy  -  98  /  -  140  yz. 

12.   Uaaj  +  ff&^  +  ta;.  13-   6ax4-15 

14.  8  aVar*  -  10  anV  +  4  aW. 

15.  45  m-V^  +  90  mWp  -  75  m*np2. 

16.  28  «563c  -  84  «  W  +  98  a  W. 

17.  27  aryz2  +  135  afyV  -  81  afy4**- 


4-7]  INTEGRAL  ALGEBRAIC   FACTORS.  101 


18.  x  —  (n  +  1)  x.  19.  a2  (a  +  x)  +  a^(a  +  SB). 

20.  3te(a-l)-3(a-l).  21.  2(n  +  l)2-4(n  +  1). 

22.  a  (#  —  1)  —  x  +  1.  23.  w  (<?  —  p)  —  (_p  —  <?). 

24.  6  mn+1  —  3  mn+2  +  9  mn+3.  25.  an+1  —  a  +  an~\ 

26.  5  M+3  -  125  x  +  625  a2.  27.  2»+4  -  8  x  2"-1  +16. 

Grouping  Terms. 

7.  When  all  the  terms  of  a  given  expression  do  not  contain 
a  common  factor,  it  is  sometimes  possible  to  group  the  terms 
so  that  all  the  groups  shall  contain  a  common  factor. 

Ex.  1.   Factor  2  a  +  2  b  -f  a#  +  da?. 

Factoring  the  first  two  terms  by  themselves,  and  the  last 
two  terms  by  themselves,  we  obtain 

2(o  +  b)  +x(a  +  b)  =  (a  +  b)  (2  +  x). 
Ex.  2.     a;2  —  ic   —  a 


Ex.3. 


EXERCISES  II. 

Factor  the  following  expressions  : 

1.    am  +  an  +  6m  +  bn.  2.  ax  —  by  —  bx-{-  ay. 

3.   m2  —  am  +  bm  —  ab.  4.  x2  —  5x  —  2xy  +  W  y* 

5.   ax  +  a  +  x  -f-  1-  6-  wa  —  a  +  71  —  1. 

7.    mz  +  m  —  z  —  1.  8.  x3  —  ce2  +  x  —  1. 

9.    a;  —  y  —  a;y  +  1.  10.  1  —  3  o  —  6  -f-  3  a&. 

11.   a?  -  a2c  +  ac2  -  c3.  12.  3«4-  ar3  +  6x-  2. 

13.   3  c4  —  3  (?n  +  en2  —  n3.  14.  5  aa?  —  ex  —  5  ay  +  cy. 

15.   2  aa?  —  3  by  —  2  ay  +  3  bx.      16.  ac  —  5  ad  -f  3  be  —  15  6tL 

17.    3n3  +  nx2-6n2x-2x*.        18.  18  n2a  -  12  x  -  9  n2  +  6. 


102  ALGEBRA.  [Cn.  VI 

19.  18ax  +  30ay-9bx-15by. 

20.  20  ad  -  35  bd  -  8  ax  +  14  6a?. 

21.  24  ww  —  44  n2  —  30  ma?  +  55  nx. 

22.  12  a3&4  -  4  a264  -  4  a?W  +  12  a363. 

23.  a4  -  a3^2  +  a2n  -  an3  +  n5  -  an3. 

24.  «4-ax3  +  3a2^-2a26x2  +  2a36x-6a45. 

25.  a#  +  fry  +  cz  +  6a?  +  cy  -f-  «z  +  ca?  4-  ay  4-  &«. 

26.  ax  —  by  -{-  cz  —  bx  —  cy  —  az  —  ex  -\-  ay  +  bz. 

27.  a#  +  by  +  c^;  —  bx  —  cy  -\-  az  -f  ex  —  a^/  —  bz. 

28.  a#  +  5?/  +  cz  —  bx  +  cy  —  az  —  c»  —  ay  4-  60. 
29.   ^  +  4^-3^-12.  30.   ^-3^  +  5^-15. 
31.   x3  +  2x2  +  8a?  +  16.               32.   ^-7^-4^  +  28. 


Use  of  Type-Forms  in  Factoring. 

8.  If  an  expression  is  in  the  form  of  one  of  the  type-forms 
considered  in  Ch.V.,  or  if  it  can  be  reduced  to  such  a  form, 
its  factors  can  be  written  by  inspection. 

Trinomial  Type-Forms. 

9.  From  Ch.  V.,  Arts.  2  and  3,  we  have 


Therefore  a  trinomial  which  is  the  square  of  a  binomial 
must  satisfy  the  following  conditions: 

(i.)  One  term  of  the  trinomial  is  the  square  of  the  first  term  of 
the  binomial. 

(ii.)  A  second  term  of  the  trinomial  is  the  square  of  the  second 
term  of  the  binomial. 

(ui.)  The  remaining  term  of  the  trinomial  is  twice  the  product 
of  the  two  terms  of  the  binomial. 


7-10]  INTEGRAL  ALGEBRAIC   FACTORS.  103 

10.   Ex.  1.   Factor  x2  4-  6  x  +  9. 

x2  is  the  square  of  x,  9  is  the  square  of  3,  and  6  x  =  2  •  x  -  3, 

Therefore  x2  +  6  x  +  9  =  (x  +  3)2. 

Ex.  2.   Factor  —  4  xy  +  4  x2  4-  y2. 

4  #2  is  the  square  of  2  x,  or  of  —  2  x ;  y2  is  the  square  of  y, 
or  of  —  y.  Since  the  term  —  4  xy  is  negative,  one  term  of  the 
binomial  is  negative,  the  other  positive. 

Therefore  -  4  xy  4-  4  x2  +  y2  =  (2  x  -  y)2  =  (-  2  x  +  y)2. 

EXERCISES   III. 

Factor  the  following  expressions : 

3.   y2  4-  12  y  4-  36.  4.   a2  — 

5.   4  cc2  —  12  xy  4-  9  y2.  6.   9  a2  - 

7.   20aj-4a?2-25.  8.   36 a?  -  4 aj8  -  81. 

9.   16  a2  4-  40  a&  4~  25  62.  10. 

11.   a4  —  2  a2x  4-  x2.  12. 

13.   4  CM?  4-  2  a2  4-  2  x2.  14. 

15.    a2a?2  —  4  ac3o?  4-  4  c6.  16. 

17.    24  ay  -  9  a2- 16  y2.  18. 

19.   4arn-12xn4-9.  20.    36  aK+2  -  48  an  4- 16  an~2. 

21.   4  a4&2  —  12  a?bc2  4-  9  c4.  22.    25  m4n4  —  60  raV^2  4-  36  p*. 

23.   16#6y4-24aryV4-9z6.  24.   49  a466  4-  70  aW  +  25  c8. 

25.  (a4-a)24-2(a  +  xN)4-l. 

26.  (2a-9)2- 

27.    xy  —  xz—  (y2  —  2yz  +  z2).    28.    a2  4-  2  an  4-  n2  —  ap  —  pn. 

31.  x2  —  6yz  —  4cxy  +  3  xz  -\-  4:  y2. 

32.  a464  4-  2  a?b3  4-  2  a262  +  2  a6  4- 1. 


104  ALGEBRA.  [Cn.  VI 

11.  From  Ch.  V.,  Art.  7,  we  have 

x-  +  (a  +  b)  x  +  ab  =  (x  +  a)  (jr  +  6). 

"When  a  trinomial,  arranged  to  descending  powers  of  some 
letter,  say  x,  can  be  factored  into  two  binomials,  it  must  satisfy 
the  following  conditions : 

(i.)  One  term  of  the  trinomial  is  the  square  of  the  letter  of 
arrangement,  i.e.,  of  the  common  first  term  of  the  binomial  factors. 

(ii.)  The  coefficient  of  the  first  power  of  the  letter  of  arrange- 
ment in  the  trinomial  is  the  algebraic  sum  of  tivo  numbers  whose 
product  is  the  remaining  term  of  the  trinomial. 

(iii.)  These  two  numbers  are  the  second  terms  of  the  binomial 
factors. 

12.  Ex.  1.   Factor  x-  +  8  x  +  15. 

The  common  first  term  of  the  binomial  factors  is  evidently 
x.  The  second  terms  are  two  numbers  whose  product  is  15, 
and  whose  sum  is  8.  Byv  inspection  we  see  that 

3  +  5  =  8  and  3  x  5  =  15; 

that  is,  the  second  terms  of  the  binomial  factors  are  3  and  5. 
Consequently,  x2  -f-  8  x  +  15  =  (a;  +  3)  (x  +  5). 

Ex.  2.   Factor  x2  -  7  x  + 12. 

The  common  first  term  of  the  binomial  factors  is  x.  The 
second  terms  are  two  numbers  whose  product  is  12,  and  whose 
sum  is  —  7.  Since  their  product  is  positive,  they  must  be  both 
positive  or  both  negative;  and  since  their  sum  is  negative,  they 
must  be  both  negative. 

The  possible  pairs  of  negative  factors  of  12  are :  —  1  and 
-12;  -2  and  -6;  -3  and  -4. 

But  since  -  3  +  (-  4)  =  -  7, 

the  second  terms  of  the  binomial  factors  are  —  3  and  —  4. 
Consequently,  x2  -  1  x  +  12  =  (x  -  3) (a?  -  4). 


11-12]  INTEGRAL  ALGEBRAIC   FACTORS.  105 

Ex.  3.    Factor  a2x*  +  5 ax-  24. 

The  common  first  term  of  the  binomial  factors  is  ax.  The 
second  terms  are  two  numbers  whose  product  is  —  24,  and 
whose  sum  is  5.  Since  their  product  is  negative,  one  must  be 
positive  and  the  other  negative;  and  since  their  sum  is  posi- 
tive, the  positive  number  must  have  the  greater  absolute  value. 
The  possible  pairs  of  factors  of  —  24  are :  —  1  and  24 ;  —  2 
and  12 ;  -  3  and  8 ;  -  4  and  6. 

But  since  —3  +  8  =  5, 

the  second  terms  of  the  binomial  factors  are  —  3  and  8. 
Consequently  aV  +  5  ax  —  24  =  (ax  —  3)  (aaj  +  8). 

Ex.  4.   Factor  x2  —  3  xy  —  28  y\ 

The  common  first  term  of  the  binomial  factors  is  x.  The 
second  terms  are  two  numbers  whose  product  is  —  28y2,  and 
whose  sum  is  —  3y.  It  is  evident  that  both  of  these  terms 
contain  y  as  a  factor.  Therefore  we  have  only  to  find  their 
numerical  coefficients. 

Since  their  product  is  negative,  one  must  be  positive  and  the 
other  negative ;  and  since  their  sum  is  negative,  the  negative 
number  must  have  the  greater  absolute  value.  The  possible 
pairs  of  factors  of  -28  are  :  1  and  -28 ;  2  and  -14 ;  4  and  -7. 

But  since  4  +  (-  7)  =  -  3, 

the  second  terms  of  the  binomial  factors  are  4y  and  —ly. 
Consequently,  x2  —  3  xy  —  28  y-  =  (x  +  4  y)  (x  —  1  y). 

EXERCISES  IV. 

Factor  the  following  expressions  : 

1.  x2  -  3  x  +  2.             2.  x2  +  3  x  +2.  3.  x2  -  x  -  2. 

4.  x2  +  x-2.                5.  x*  +  x-6.  6.  x2-x-6. 

7.  x2  +  Ix  +  6.             8.  a?  -  5x  +  6.  9.  x2  +  lOz  -  24. 

10.  x2  -  2x  -  24.         11.  x2  +  5z  -  24.  12.  x2  -  23x  -  24. 

13.  x2  -  5x  -  24.         14.  x2  +  23 a;  -  24.  15.  x2  +  2x  -  24. 


106  ALGEBRA.  [Cn.  VI 

16.  x>  -  lOx  -  24.       17.  a?  +  3x  -  40.       18.  x2  -  18  x  -  40. 

19.  x2  +  6  #-40.         20.  x2--39x-40.     21.  x2-  4x-60. 

22.  #2  +  7x-30.         23.  x2  +  12x  +  32.     24.  a2-3x-40. 

25.  x2  -  12  a;  +  35.       26.  x3-17  x2+72  x.    27.  x2  +  13  x  -  30. 

28.  ex-x^x3.          29.  35  +  2x-x2.        30.  a;4 +  4 a2  — 21. 

31.  x4  +  8#2  +  15.        32.  x4-24x2  +  63.     33.  3x6+39x3+66. 

34.  x6  -  x3  -  56.  35.  a*»+6  x"-112.     36.  x2"  -  16  xn  +  55. 

37.  a?2  +  (a  -f  &)  a?  +  a&.  38.  cc2  —  (m  +  w)  x  +  wi»i. 

39.  ic2  +  (_p  —  q)x  —  pq.  40.  x2  +  (3  r  —  2  s)  a?  -  6  rs.    • 

41.  a^  +  7a2«  +  6a3.  42.  x2  +  2xy-15y2. 

43.  #2 -4  ax -12  a2.  44.  ic2  -  7  ax  +  12  a2. 

45.  2xy-26xy  +  84x2/4.  46.  or  -  11  xm  +  30  m2. 

47.  x2z2  +  12xz-  13.  48.  a2b2  -7ab  +  10. 

49.  m2n2  -  20  mn  +  99.  50.  1  -  25  ay  +  126  or0/. 

51.  !-23a26  +  132a462.  52.  oV  -  23  a2x  +  120. 

53.  ajy  -7x2y2-  78.  54.  a466  +  3  as6»  -  108. 

55.  a468  +  5  a264^  -  84  x4.  56.   a2"62n  -  2  an&"c?  -  15  c4. 

13.  From  Ch.  V.,  Art.  8,  we  have 

(fljr  +  6)  (ex  +  (/)  =  act2  +  (at/  4-  6c) JT  +  M. 

A  trinomial  which  can  be  factored  by  this  type-form  must 
satisfy  the  following  conditions  : 

(i.)  One  term  of  the  trinomial  is  the  product  of  the  first  terms 
of  its  binomial  factors. 

(ii.)  A  second  term  of  the  trinomial  is  the  product  of  the  second 
terms  of  its  binomial  factors. 

(iii.)  The  remaining  term  of  the  trinomial  is  the  sum  of  the 
cross-products. 

Ex.  1.    Factor  6  x2  +  19  x  +  10. 

The  first  terms  of  the  required  binomial  factors  are  factors 
of  6  x2,  the  second  terms  are  factors  of  10,  and  the  sum  of  the 
cross-products  is  19  x. 


12-13]  INTEGRAL   ALGEBRAIC   FACTORS.  107 

The  factors  of  6  x2  are :  x  and  6  x,  2  x  and  3  x ;  and  the  factors 
of  10  are :  1  and  10,  2  and  5. 

The   following   arrangements    represent    possible   pairs   of 
factors : 

a-  +  1  a?  +  10  x  +  2 


6x  + 


h. 


x 


16  a?  61a;  17  x 

2#  +  10  2x  +  2 


XXX 

&  +  1  3».+  5  3z  +  2 


32x  16x  19x 

Since  the  sum  of  the  cross-products  in  the  last  arrangement 
is  equal  to  the  middle  term  of  the  given  trinomial,  we  have 

6  or2  +  1 9  a  +  1 0  =  (2  a  +  5)  (3  a  +  2) . 

Ex.  2.   Factor  5  x2  -  6  xy  -  8  y2. 

The  factors  of  5  x2  are  x  and  5  x ;  and  the  factors  of  —  8  yz 
are  :  y  and  —  8  y,  —y  and  8y,2y  and  —  4  y,  —2y  and  4  y. 
x  —  2y          Since  the  sum  of  the  cross-products  in  the  arrange- 
ment on  the  left  is  equal  to  the  middle  term  of  the 
given  trinomial,  we  have 


Ex.  3.   Factor  10  a4  +  a?b  -  21 62. 

The  factors  of  10 a4  are:  a2  and  10 a2,  2 a2  and  5 a2;  and  the 
factors  of  -  21 W  are :  b  and  -  21  6,  -  b  and  21  6,  3  b  and  -  7  6, 
-  3  b  and  7  b. 

2  a2  +  3  b        since  the  sum  of  the  cross-products  in  the  arrange- 

Xment  on  the  left  is  equal  to  the  middle  term  of  the 
-2      _  ,     given  trinomial,  we  have 
~tfb~  10a4  +  a26  -  21  b2  =  (2  a2  +  3  b)  (5  a2  -  76). 


108  ALGEBRA.  [CH.  VI 

14.  The  following  directions  may  be  observed  in  factoring 
trinomials  which  come  under  this  type-form  : 

(i.)  When  all  the  terms  of  the  trinomial  are  positive,  only  posi- 
tive factors  of  the  last  term  are  to  be  tried. 

(ii.)  When  the  middle  term  is  negative  and  the  last  term  is 
positive,  the  factors  of  the  last  term  must  be  both  negative. 

(iii.)  When  the  middle  term  and  the  last  term  are  both  negative, 
one  factor  of  the  last  term  must  be  positive,  the  other  negative. 

(iv.)  /Select  those  pairs  of  factors  of  the  jirst  and  last  terms 
which,  by  cross-multiplication,  give  the  middle  term  of  the  trinomial. 

EXERCISES  V. 
Factor  the  following  expressions : 

1.   60?  + a;  — 12.  2.    6x*-x-l2. 

3.   35x2  +  32x-12.  4.   35  a;8  +  a? -12. 

5.   35  a;2  +  16  a;  —  12.  6.   35^  — 

7.   2a?2  +  5a;  +  2.  8.   10 

9.   6  +  13^-63^.  10.   3^  +  13^  +  12. 

11.   40  +  2^-2^.  12.    25or5  +  25.T2-6#. 

13.   36 x4-  18 a2 -10.  14.    12  a;  —  6  a;2  —  90  a?. 

15.  10x2+Jx-33.  16.   8^-19^-15. 
17.   40  +  6Z-2702.  18.   49£2-35x  +  6. 
19.    64^-92^  +  30.  20.    6- 

21.    6^-41^-56.  22.   30^-89^ 

23.   18x2-3xy-±5y\  24.   3a2-5ab-2b2. 

25.    abx2-(a2  +  b2)x  +  ab.       26.   abx2  +  (a2  -  b'2) x  -  ab. 

27.    5aV-4a2xz-96z2.        28.    -  10  a4  +  7  a2b2  +  12  b4. 

29.   4x*  —  xy-3y2.  30.    10a2  +  11  ab  -  6b2. 

31.    9ar2n-4af-5.  32.    2  x*r+2  -  3  af+1  -  2. 

33.   6x2m  +  xmyn-Wy2n.  34.   10(a  +  &)2+7c(a+6)-6c2. 

35.  7(x-y)*-37z(x-y)  +  Wz>. 

36.  6  (x2  +  y*f  -  9  (x2  +  f)  z2  -  15  24. 

37.  2(a2-c2)2-46(a2-c2)-662. 


14-16]  INTEGRAL   ALGEBRAIC   FACTORS.  109 

Binomial   Type-Forms. 

15.  From  Ch.  V.,  Art.  6,  we  have 

a2  —  b2  =  (a  +  b)  (a  —  b). 

That  is,  the  difference  of  the  squares  of  two  numbers  can  be 
written  as  the  product  of  the  sum  and  the  difference  of  the 
numbers. 

Ex.  1.  a?x-  -  1  ft2  =  (ax)2  -  (J  ft)2 

=  (ax  -f-  -J-  ft)  (ax  —  -J-  ft). 

Ex.  2.  32  m*n  -2n3  =  2  n  (16  m4  -  n2) 

=  2n[(4m2)2-n2] 
=  2  n  (4  m2  -f  w)  (4  m2  -  n). 

16.  The  difference  of  any  even  powers  of  two  numbers  can 
be  written  as  the  difference  of  the  squares  of  two  numbers,  and 
should  therefore  first  be  factored  by  applying  this  type-form. 

Ex.  a4  -  ft4  =  (a2)2  -  (ft2)2 

=  (a2  +  ft2)  (a2  -  ft2) 


EXERCISES  VI. 

Factor  the  following  expressions  : 


1. 

X2 

-1. 

2. 

4  -a2. 

3. 

16  -  y2. 

4. 

25 

x2y2 

— 

9. 

5. 

36  a2  -  49  ft2. 

6. 

&X2  —  y 

|4 

7. 

86 

2  -  142. 

8. 

572  -  432. 

9. 

372  _  2' 

10. 

81 

a4- 

-16. 

11. 

4  a262  _  |5  C2d: 

\       12. 

16  a6  - 

25  ft4c6. 

13. 

o?l 

AJo  _ 

~  i 

14. 

iaV-Ti__x( 

5.        15. 

a2"  -  1. 

16. 

a2' 

17. 

x2n+2  —  4. 

18. 

9  a2nft2  - 

-4C2"*. 

19. 

7- 

-11 

2  x 

4 

20. 

16  a4  -  f. 

21. 

a8  -  ft8. 

22. 

1  -  256  afy8 

23. 

x™  -  2/16. 

24. 

a16-!. 

25. 

5a2- 

180  ft4. 

26. 

2  a62  _  1  a(,2> 

27. 

5  XV*  • 

-5.CCZ6. 

28.   75a2ft4-108c2d4.  29.   243  ft5c6  -  75  ft7. 

30.   a4*—  ft4*.  31.   144 a"  —  a;w+2.         32. 


110  ALGEBRA.  [Cn.  VI 

33.    a2  -  b2  -f  (a  +  b)c.  34.    a2  -  x2  +  a  -  x. 

35.    a4  —  a3  -f  a  —  1.  36.   x2  —  xz  —  yz  —  if. 

37.    a2  -  a?n  +  an2  -  n2.  38.    a4  -  2  a&3  -  64  +  2  a36. 

39.    a?y  —  xf  +  xiy  +  xy2.  40.    a2  +•  3  or5  -  a4  -  3  a. 

41.  (a  -f-  n)  (a2  -  x2)  -  (a  -  x)  (a2  -  n2). 

42.  (n  -  x)  (5n2  -  4x-2)  -  (3  x2  -  4  ti2)  (a  -  n). 

17.   This   type-form   may  frequently  be   applied  to  multi- 
nomials. 

Ex.  1.     ^- 


Ex.  2. 

=  (2ac  +  a2  -  b2  +  c2)  (2ac  -  a2  +  62  -  c2) 


=  (a  +  c  +  6)  (a  +  c  —  6)  (6  +  a  —  c)  (6  —  a  +  c). 

EXERCISES   VII. 

Factor  the  following  expressions  : 

1.    (a  +  6)2-c2.          2.    (a-6)2-c2.         3.    (n  +  I)2  -  n2. 

4.   n2-(w-l)8.          5.   9-(3-z)2.          6.    49-4(a  +  5)2. 

7.    (2  a  -f-  6)2  -  9  c2.  8.    (4^-3)2-16a^. 

9.    2oa2-4(6  +  c)2.  10.    36  'X2  -  81  (x  -  2)2. 

11.    (a  +  6)2-(c  +  cZ)2.  12.    (a-&)2-(c-d)2. 

13.    (a  +  Z>)2-(a-&)2.  14.    (»  +  2)2-(a?-l)2. 

15.    (5x-2)2-(4z-3)2.  16.    (3  ^-4)2-  (2^-6)2. 

17.    (a  +  6-c)2-(a-6  +  c)2.    18.    (x+y-3)2-(x-y+5)2. 

19.  (x2  +  x-fl)2-(^ 

20.  (a  +  &)2-l-2(a+6 

21.  (a-26)2-9-3(a- 

22.    ^-2x2/4-?/2-z2.  23.    a~-2 

24.    z2  —  x2  —  2xy  —  y2.  25.    9  —  or2  -f-  2  xy  —  y2. 


16-18]  INTEGRAL   ALGEBRAIC   FACTORS.  Ill 

26.   a2-n2  +  2np-p2.  27.    a2  +  2  be  -  b2  -  c2. 

28.   25  +  12a2/-9^-42/2.         29.   25Z2  -  49?/2-  10  a?  +  1. 

30.  a2  -  2  a&  +  b2  -  x2  -  2  xy  -  y2. 

31.  x2- 

32.  aa  + 

33.  a2  + 

34.  25x2 

35.  a4- 

36.  4a4 

37.  a2  + 

38.  a2  +  62-c2-d2-2a6-cd). 

39.  2  (06  +  cd)  -  (a2  +  b2  -  c2  -  d2). 

40.  a2  -  b2  +  2  fo  -  2  aa  +  aj2  -  z2. 

41.  4a262-(a2  +  62-c2)2. 

42.  a2r  -  a47"  -  2  a7r  -  a10r. 

43.  a4  +  4a2c-462  +  45d2  +  4c2- 

44.  4  (ad  +  6c)2  -  (a2  -  b2  -  c2  +  d2)2. 

18.   From  Ch.  V.,  Art.  10,  we  derive 


Ex.1. 


Ex.  2.  512  x*-ys  = 

= 

=  (8  x2  -  y)  (64  cc4  +  8  x2y  +  2/2) 
Ex.  3. 

a6  -  729  66  =  (a3)2  -  (27  63)2 

=  (a3  +  2763)(a3-2763) 

=  (a+3  6)(a2-3  a6+9  fe2)  (a-3  6)(a2 


112  ALGEBRA.  [Cn.  VI 

Ex.  4. 

(1  -  a;)8  -  8  a8  =  (1  -  x)*  -  (2  xf 


19.  The  sum  of  the  like  even  powers  of  two  numbers,  whose 
exponents  are  divisible  by  an  odd  number,  except  1,  can  be 
factored  by  applying  the  type-forms  of  Art.  18. 

Ex.  aP       "  =  i* 


EXERCISES   VIII. 
Factor  the  following  expressions  : 

l.  £3  +  l.  2.   a?  -8.  3.   a3  +  27. 

4.  64  ar3-!.  5.    8  or5-/.  6.   8^-27. 

7.  125  ofy6  +  8.         8.    3  a2  -24  a5.  9.   27a-a46fi. 

10.  27  a5-/.  11.    125  ar5-?/12*12.         12.    2arV  +  432/. 

13.  27aW+l.      14.    64zyz9-125.       15.   8wV-343p9. 

16.  a6  -64.  17.    x6  +  2/6.  18.    X9  +  /. 

19.  .T9-l.  20.    a12-!.  21.    a12  +  612. 
22.  1-218.                  23.    a18  +  ^18.  24.    a3"  -ft3". 
25.  8  ar*tym  -  729  tT+V.                    26.    (a;  +  y)8-l. 

27.  l_(aj-^)8.  28.    27  —  (3  +  2  a?)8. 

29.  (a  +  by  4-  (a  -  6)8.  30.    (2  x  -  I)3  -  (a:  -  2)3. 

31.  (2a4-^)3  +  (a-2«)8.  32.    (a  4-  ^)3  -  (c  +  d)3. 

33.  x*  —  y*  —  2xry  +  2xy2.  34.    4  —  ^  +  4^  —  x5. 

35.  ar5  —  a?1  -  a;2  -f  1.  36.    a,-8  -  8  -  Qx2  +  12  a?. 

37.  a3  -  4  a"c  —  4  ac2  H-  c8.  38.    ?i6  +  5  71  V  +  5  ?i  V  4-  a£ 

20.  From  Ch.  V.,  Art.  11,  we  derive  : 

(i.)    7%e  swm  o/  the  like  odd  jwwers  of  two  numbers  contains 
the  sum  of  the  numbers  as  a  factor. 


18^22]  INTEGRAL   ALGEBRAIC   FACTORS. 

(ii.)    The  difference  of  the  like  odd  powers  of  two  numbers  con- 
tains the  difference  of  the  numbers  as  a  factor. 

Ex.  1.    x>  +  y5=  (x  +  y)  (x4  -  tfy  +  X2y2  -  xf  +  y*). 
Ex.  2.    x7  -  y7=  (x  - 


EXERCISES   IX. 

Factor  the  following  expressions  : 
1.    a5  +  &5.  2.    x>  —  1.  3. 

4.    a7-!.  5.   32a-5-610.  6.   243  a10-?/5. 

7.    a10  +  fe10.  8.    x10-!.  9.    #15  +  1. 

10.    128x7  +  l.  11.    a565      32.  12.    ar'10  -  1024  »1 


Special  Devices  for  Factoring. 

21.  A  factorable  expression  can  frequently  be  brought  to 
some  known  type-form  by  adding  to  or  subtracting  from  it  one 
or  more  terms. 

Ex.  l.    Factor  x*  +  x2y2  +  y\ 

This  expression  would  be  the  square  of  a^  +  y2,  if  the  co- 
efficient of  x~y2  were  2.  We  therefore  add  x2y2  ;  and,  in  order 
that  the  value  of  the  expression  may  remain  the  same,  we- 
subtract  x2y2.  We  then  have 

x*  +  2  x2y2  +  if-  x2y2=(x2  +  y2)2  -  x2y2 


22.  Another  device  consists  in  separating  a  term  into  two 
or  more  terms,  and  grouping  these  component  terms  with 
others  of  the  given  expression. 

Ex.   Factor  x3  -  3  x2  +  4. 

Separating  —  3  x2  into  —  2  x2  and  —  ic2,  we  obtain 


=  (x  —  2)(x2-x-2) 


114  ALGEBRA.  [On.  VI 

EXERCISES   X. 

Factor  the  following  expressions  : 
1.    l  +  4a4.  2.    1  +  64  a4.  3.   a4" +  4 1/4". 

4.  l  +  3a2  +  4a4.      5.   l-7a2  +  a4.         6. 

7.    aj'-ay+lG  y\       8.    a4+2/4-llxy.       9.    16  x*- 

10.  x4  +  4  2T4  -  12  ajy.  11.    x4  +  2/8  +  tf2?/4. 

12.  x8  +  y8  -  142  afy* .  13.   or5 -6  ^  +  16. 

14.  ^-15^  +  250.  15.    or?  +  6.T2  +  10.T  +  4. 

16.  0^-9  ^4- 32  a? -42.  17.    jc8  -  15  a;8  +  72  a?  -  110. 

18.    8  or3  -36  or2  +  48  aj  -  18. 

EXERCISES  XI. 

Factor  the  following  expressions  by  the  methods  given  in 
this  chapter : 

1.  a4  +  2a36-2a&3-&4.  2.    ax2  +  (a  +  b  +  c)x  +  &  +  c. 

3.  10c4n+1-5c7"+1-5cn+1.  4.   «2?/2  +  17  xy  +  16. 

5.  x6  +  64.          6.    a666  +  l.  7.    2336+s  — 64.  8. 
9.  2a4-16a&».  10.   aj*  +  2a;2  +  9. 

11.  24a?B-(3&-8a)a;-a*.    12.    62-  c2  +  a  (a  -  26). 

13.  x-2m-2  +  2 am+n  +  a;2n+2.          14.    x4  -  2 a?-  1  +  2x. 

15.  or +  11  a +  24.  16.    a2-a6-662. 

17.  tff  —  4  xy  —  5.  18.    ic2  +  x  +  y  —  y2. 

19.  aft  (x2  +  if)  +  spy  (a2  +  &2). 

20.  28  (x  +  3)2  -  23  (x2  -  9)  -  15  (a?  -  3)2. 

21.  aa?5  +  6a^  +  c,y?  —  ax2  —  bx  —  c. 

22.  (a  +  6)  y?  +  (a  -  2  6)  x  -  3  b. 

23.  a2  -  62  -  c2  -  2  a  +  2  6c  +  1. 

24.  49  x\f  +  42  z7?/9  +  9  ^y 2. 

25.    x2-13^/  +  40/.  26.    a2-5a&  +  6&2. 

27.    m2n2  +  6  ?/i7i  —  55.  28.    &2  +  ac  —  c2  +  a&. 


22]  INTEGRAL   ALGEBRAIC   FACTORS.  115 

29.    xy-xz-\-2yz-y2-z2.  30.    x2  -  2x  +  1  -  ?/2. 

31.    15  a2  +  #  —  40.  32.    x3  —  x2z  -f-  #z2  —  z3. 

33.    a3  _  1  _j_  c  _  ac.  34.    a2  _  a  —  1  —  a?c  +  ac  +  c. 

35.   2a2  +  a-4:ax-x  +  2x2.        36.   20^-123^  +  180. 

37.   x3-5x2-x  +  5.  38.   «2  (a +  1)- 62  (6  +  1). 

39.   25  a464  +  70  aW  +  49  c4.         40.    x4y  +  zx*  -  xy  -  z. 

41.  ^-9^2-4?/(2/-f  3z). 

42.  a?8  -  2  ajy  +  /  -  4  a?y  («2  -  ?/2)2. 

43.   a3  +  a2c  +  a&c  +  52c -  63.          44.   5  a4- 10  a3  -  75  a2. 

45.   3(a-l)3-(l-a).  46. 

47.   «2 -ax -bx  +  db.  48. 

49.   a;2  +  9-2x(3  +  2^2).  50. 

51.   3xQ  +  8^-8^-3.  52. 

53.    (tf+xy+y^-^-xy+y*)2.     54. 

55.    (a^  +  I)3  -  (y2  +  I)3.  56.    aftaj8  +  a;  +  ab  +  1. 

57.   36a4-21a2  +  l.  58.   10  x4  -  47  x2  +  42. 

59.  (x2  +  ^-2/2)2-(^-^-2/2)2. 

60.  x2  +  c  (a  +  6)  ic  +  a6  (a  +  c)  (c  —  6). 

61.   5 a2-  180 b2.  62.   ^ ab<? - -£r abd2. 

63.   10^  +  3^-18.  64.   x2n-/n-f  42/n- 

65.   a6(^-2/2)  +  ^(a2-62).          66.   36  a462  -  60  a;3&3  + 

67.   a2(a2-l)-62(62-l).  68.    (m-n)2-12(ra-?i)+27. 

69.  a2xs(a3-x)-a5x2(x3-a). 

70.  (a2-62)(a  +  6)  +  2a62-2a26. 

71.  (a  -  6)2  -  x2  -  (x  -  a  +  6)  (a  +  &  -  x). 

72.  (a +  2/)2- 18  <>  +  ?/)  + 77. 

73.  (a2  -  62)  or2  -  (a2  +  62)  a  +  ab. 

74.   300a&c2-432a&d2.  75.   75  a262  -  108  c2^2. 

76.   ^abx2y2-^9-abz2.  77.    18  aV  -  98  &y. 

78.   18  (x  +  t/)2  +  23  (x2  -  y2)  -6(x-  y)2. 
79.   Express  (a2  —  b2)  (c2  —  d2)  as  the  difference  of  two  squares. 


116  ALGEBRA.  [Cn.  VI 

HIGHEST   COMMON  FACTORS. 

23.  If  two  or  more  integral  algebraic  expressions  have  no 
common  factor  except  1,  they  are  said  to  be  prime  to  one 
another. 

E.g.,  ab  and  cd ;  5  tfy  and  8  z8 ;  a2  +  b2  and  a2  -  b'2. 

24.  The  Highest  Common  Factor  (H.  C.  F.)  of  two  or  more 
integral   algebraic  expressions   is   the   expression  of  highest 
degree  which  exactly  divides  each  of  them. 

E.g.,  the  H.  C.  F.  of  ax2,  ba?,  and  ex4  is  evidently  a?. 

25.  Monomial  Expressions.  —  The  H.  C.  F.  of  monomials  can 
be  found  by  inspection. 

Ex.  1.    Find  the  H.  C.  F.  of  x2y5z,  x4y*z2,  and  afyV. 

In  the  expression  of  highest  degree  which  exactly  divides 
each  of  the  given  expressions,  the  highest  power  of  x  is  evi- 
dently x2,  of  y  is  y*,  and  of  z  is  z.  Therefore  the  required 
H.  C.  F.  is  x2ysz. 

Observe  that  the  power  of  each  letter  in  the  H.  C.  F.  is  the 
loivest  power  to  which  it  occurs  in  any  of  the  given  expressions. 

If  the  monomials  contain  numerical  factors,  the  Greatest 
Common  Measure  (G.  C.  M.)  of  these  factors  should  be  found 
as  in  Arithmetic. 

Ex.  2.    Find  the  H.  C.  F.  of  18  aWd,  42  a?bc4,  and  30  aW. 

The  G.  C.  M.  of  the  numerical  coefficients  is  6.  The  lowest 
power  of  a  in  any  of  the  given  expressions  is  a2 ;  the  lowest 
power  of  b  is  b ;  the  lowest  power  of  c  is  c2 ;  and  d  is  not  a 
common  factor.  Therefore  the  required  H.  C.  F.  is  6  arbc2. 

26.  In    general,   to    obtain    the   H.  C.  F.    of   two    or   more 
monomials : 

Multiply  the  G.  C.  M.  of  their  numerical  coefficients  by  the 
product  of  their  common  literal  factors,  each  to  the  lowest  power 
to  which  it  occurs  in  any  of  the  given  monomials. 

27.  Multinomial  Expressions.  —  The  method  of  finding  the 
H.  C.  F.  of  multinomials  by  factoring  is  similar  to  that  of  find- 
ing the  H.  C.  F.  of  monomials. 


23-27]  HIGHEST   COMMON    FACTORS.  117 

• 

Ex.  l.    The  expressions 


and  x2  +  x  -  2  =  (x  -  1)  (x  +  2), 

have  only  the  common  factor  x  —  1.     This  is  their  H.  C.  F. 

In  general,  the  H.  C.  F.  of  two  or  more  multinomial  expres- 
sions is  the  product  of  their  common  factors,  each  to  the  lowest 
power  to  which  it  occurs  in  any  of  them. 

Ex.  2.  Find  the  H.  C.  F.  of  a2x2  —  a2,  2  ax2  +  2  ax  —  4  a,  and 
4  ax2  —  12  ax  4-  8  a. 

We  have  a?x*  —  a2  —  a2  (x  +  1)  (a;  —  1), 

2  ax2  +  2  ax  -  4  a  ==  2  a  (a;  +  2)  (a?  -  1), 
4  ax2  -  12  ax  +  8  a  =  4  a  (a;  -  2)  (x  -  1). 
Therefore  the  required  H.  C.  F.  is  a  (x  —  1). 

EXERCISES  XII. 

Find  the  H.  C.  F.  of  each  of  the  following  expressions  r 

1.    36  a2,  27  a4.  2.    20  ab2,  35a26.  3.    45ary,  12afys. 

4.    a2bx\  aWx2,  absx\  5.    56a?y,  70  x2/,  .  98  afy2  . 

6.    24a26a?*,  42  az3,  18a8afy.       7.    15  mV/,  40mVa;,  35m3nx2. 

8.    9(a?  +  y),  6(a-h2/)2.  9.    12y2(a-b),  30  y  (a  —  b)2. 

10.    £2-9,  x2  +  3x.  11.   3^-3^,  5aj-5ajy*. 

12.    (a  +  6)2,  a2-62.  13.    a^2  -  a,  ax2  +  2  ax  -f  a. 

14.    x2  -  25  /,  x2  +  a^  -  30.     15.    (a26  -  a&2)2,  a6  (a2  -  62). 
16.    27a^  +  /,  9  or2-/.  17.    a3  -  4  ab2,  a*-8bs. 

18.  x2-2i»-15,  «2  +  10  a?  +  21. 

19.  oj2-2a?-24,  aj2  +  9a;  +  20. 

20.  3x3-32/3,  x2-by  +  bx-xy. 

21.  «3-/,  x4  +  3^LV2-42/4. 

22.  or  +  z?/-30/,  ^2-2^-15/. 

23.  x2y2-xys-A2y4,  6  orfy  +  18  a;2/  -  108  a#* 
24. 

25. 


118  ALGEBRA.  [Cn.  VI 

• 

26.  a3-f-2a2  +  2a  +  l,  a3  -f  1. 

27.  x2  4-  a&  —  ax  —  ~bx,  x2  —  ab  —  ax  +  60;. 

28.  a2  -  (6  -  c)2,  (a  -  c)2  -  b2. 

29.  a?3  —  y3,  #4  +  a?2?/2  -f-  y4. 

30.  #2-3^  #2-9,  x-2-6z  +  9. 

31.  a8 -8,  o'  +  Taj-lS,  ^-8x  +  12. 

32.  a2-3x-40,  z2  +  3x-10,  ^-oj-SO. 
•33.  x2  -\-  2  xy  -\-  y2  —  z2,  ax  +  ay  -f-  ^z« 

.34.    (y  -  z)2  -x2,  (x  +  y)2  -  z2,  y2  -  (z  -  *)2. 

LOWEST  COMMON   MULTIPLES. 

28.  A  Multiple  of  an  integral  algebraic  expression  is  an  ex- 
pression which  is  exactly  divisible  by  the  given  one. 

E.g.,  multiples  of  a  +  b  are  2  (a  +  5),  (a?  —  y)  (a  +  6),  etc. 

29.  The  Lowest  Common  Multiple  (L.  C.  M.)  of  two  or  more 
integral  algebraic  expressions  is  the   integral   expression   of 
lowest  degree  which  is  exactly  divisible  by  each  of  them. 

E.g.,  the  L.  C.  M.  of  aa?2,  bx3,  and  ex4  is  evidently  abcx*. 

30.  Ex.  1.    Find  the  L.  C.  M.  of  a*b,  a?bc?,  and  «62c4. 

In  the  expression  of  lowest  degree  which  is  exactly  divisible 
by  each  of  the  given  expressions,  the  lowest  power  of  a  is  evi- 
dently a3,  of  b  is  b2,  and  of  c  is  c4.  Therefore  their  L.  C.  M.  is 

<X362C4. 

Observe  that  the  power  of  each  letter  in  the  L.  C.  M.  is  the 
highest  power  to  which  it  occurs  in  any  of  the  given  expres- 
sions. If  the  expressions  contain  numerical  factors,  the 
L.  C.  M.  of  these  factors  should  be  found  as  in  Arithmetic. 

Ex.  2.    Find  the  L.  C.  M.  of 

3a&2,  66(a  +  y)2,  and  4  a2b  (x  -  y)  (x  +  y). 
The  L.  C.  M.  of  the  numerical  coefficients  is  12. 
The  highest  power  of  a  in  any  of  the  expressions  is  a2 :  of 
b  is  b2 ;  of  x  +  y  is  (x  -f  y)2 ;  and  of  x  —  y  is  x  —  y. 

Consequently  the  required  L.  C.  M.  is  12  a*b'2  (x  +  y)2  (x  —  y). 


27-31]  LOWEST   COMMON  MULTIPLES. 


31.  In  general,  to  obtain  the  L.  C.  M.  of  two  or  more- 
monomials  : 

Multiply  the  L.  C.  M.  of  their  numerical  coefficients  by  the  prod- 
uct of  all  the  different  prime  factors  of  the  expressions,  each  to 
the  highest  poiuer  to  which  it  occurs  in  any  of  them. 

EXERCISES   XIII. 

Find  the  L.  C.  M..of  the  following  expressions  : 
1.   3  a,  56.  2.   Sxif,  Sx2y2. 

3.    8a26,  12a2c2,  10  ad.  4.    30u364,  45  a463,  72a262. 

5.   12ary,  18a*y,  36afy4.       6.   15a263,  60  oW,  726V. 

7.  40  aW,  62  a  W,  124  aW. 

8.  56m2na,  72m4n2y4,  90  mV?/3. 

9.   3  aj,  5  x2  +  10  #.  10.    6  w»i,  4  m2  —  12  mn. 

11.    ar2  -  1,  cc  +  1.  12.    3  a  -  6  6,  «2c  -  4  62c. 

13.   aj  +  1,  x2-2x-3.          14.    aaj  —  60;,  a3-2a26  +  a62, 
15.    (a  +  6)2,  a2  -  62.  16.    x2  (m  -  w),  a;  (m3  -  n3). 

17.  a^  +  3a;  —  10,  ^-3x-40. 

18.  ^  +  6^-55,^-11^  +  30. 

19.  x2-  4  ax  +  3  a2,  ar>  +  2az-3a2. 

20.  m2  +  2  win  —  15  w2,  m2  +  3  m?i  —  10  n2. 

21.  a3  —  aj3,  a2  —  x2,  «  —  a.     22.   x2  —  y2,  (x  —  i/)2,  or3  —  y3. 

23.  a?  —  a,  a2  —  x2,  x4  —  a4.     24.    1  —  2  x,  4  or2  —  1,  1  +  4  x2. 

25.  aj»  -  11  a;  +  24,  aj"-6aj-16,  x2  -  a;  -  6. 

26.  a^  —  4a?  —  45,  a?2  -7^-18,  o^  +  7^  +  10. 

27.  3^  +  24^  +  45,  6^  +  18^-60,  8^-24^  +  16. 

28.  4.^  +  4^-224,  6^  +  24^-462,  8^  +  64a;-264. 

29.  x2  -  4  ax  +  3  «2,  x2  +  4  az  -  5  a2,  a?  +  2  ax  -  15  a2. 

30.  x2  +  2  mx  -  3  m2,  or2  +  7  mx  -  8  m2,  x2  -  6  ma?  -  27  m2. 

31.  a?-4a2,  a?  +  2a®2  +  4a2a;  +  8a3,  ^  -  2  aar  +  4  a?x  -  8  a* 

32.  a-  -  (6  +  c)2,  62  -  (a  +  c)2,  c2  -  (a  +  6)2. 


120  ALGEBRA.  [Cn.  VI 

H.  C.  F.    AND   L.  C.  M.    BY  DIVISION. 

32.  If  the  given  expressions  cannot  be  readily  factored,  their 
H.  C.  F.  can  be  obtained  by  a  method  analogous  to  that  used  in 
Arithmetic  to  find  the  G-.  C.  M.  of  numbers. 

33.  The  expressions  whose  H.  C.  F.  is  required  should  be 
arranged  to  powers  of  a  common  letter  of  arrangement. 

If  one  of  two  expressions  be  divisible  without  a  remainder  by 
the  other,  which  must  be  of  the  same  or  lower  degree  in  the  letter 
'Of  arrangement,  then  the  latter  (the  divisor)  is  the  required  H.  C.  F. 

For  it  is  a  factor  of  the  other  expression. 

But  if  the  one  expression  be  not  divisible  without  a  remain- 
der by  the  other,  their  H.  C.  F.  is  found  as  follows  : 

(i.)  Divide  the  expression  of  higher  degree  in  a  common  letter 
of  arrangement  by  the  one  of  lower  degree  ;  if  the  expressions 
be  of  the  same  degree,  either  may  be  taken  as  the  first  divisor. 

(ii.)  Continue  the  division  until  the  remainder  is  of  loicer 
degree  than  the  divisor  in  the  letter  of  arrangement. 

(iii.)  Divide  the  first  divisor  by  the  first  remainder,  the  first 
remainder  (second  divisor)  by  the  second  remainder,  and  so  on, 
until  a  remainder  0  is  obtained.  The  last  divisor  ivill  be  the 
required  H.  C.  F. 

34.  Ex.   Find    the    H.  C.  F.    of    2x3-5x2-5x  +  S    and 
3.2  _  4  x  _j_  3 

We  have 


6x 


3x2-llx 
3x2-12x-\-9 


x2-    x 
-3x 
-3x 
By  Art.  33  (iii.),  the  H.  C.  F.  is  x  -  1. 


32-36]  H.C.F.   AND   L.C.M.   BY   DIVISION.  121 

35.  The  validity  of  the  preceding  method  is  based  upon 
the  following  principle : 

If  an  integral  algebraic  expression  be  divided  by  another  (of 
the  same  or  loiver  degree  in  a  common  letter  of  arrangement) 'and 
if  there  be  a  remainder,  then  the  H.  C.  F.  of  this  remainder  and 
the  divisor  is  the  H.  C.  F.  of  the  given  expressions. 

E.g.,  the  H.  C.  F.  of 
^_  10  ^+  35  X2_  50  x  +  24?  =  (<,.  _  i)  (3  _  2)  (a?  -  3)0  -  4),  (1) 

and         x3-7x2  +  llx-5,  =  (a?  - 1)  (a?  - 1)  (a?  -  5)  (2) 

is  evidently  x  —  1. 

The  remainder  obtained  by  dividing  (1)  by  (2)  is 

3o2-12a  +  9,  =3(a;-l)(a?-3).  (3) 

The  H.  C.  F.  of  this  remainder  and  the  divisor  (2)  is  evi- 
dently also  x  -  1,  the  H.  C.  F.  of  (1)  and  (2). 

Notice  that  the  H.  C.  F.  of  the  remainder  and  the  dividend 
(1)  is  (a;  —  1)  (a;  -  3),  and  is  not  the  H.  C.  F.  of  (1)  and  (2). 

Since  this  principle  can  be  applied  at  any  stage  of  the  work, 
the  H.  C.  F.  of  any  remainder  and  the  corresponding  divisor  is 
the  required  H.  C.  F. 

When  the  last  remainder  is  0,  the  last  divisor  is  the  H.  C.  F.  of 
itself  and  the  corresponding  divisor,  that  is,  of  the  preceding 
remainder  and  divisor,  and  is,  therefore,  the  required  H.  C.  F. 

If  a  remainder  which  does  not  contain  the  letter  of  arrange- 
ment, and  which  is  not  0,  is  obtained,  the  given  expressions  do 
not  have  a  H.  C.  F.  in  this  letter  of  arrangement. 

The  proof  of  the  principle  enunciated  is  given  in  School 
Algebra,  Ch.  VIII. 

36.  The  following  principle  will  frequently  simplify  the 
work  of  finding  the  H.  C.  F.  of  two  expressions : 

Either  of  the  expressions  may  be  multiplied  or  divided  by  any 
number  which  is  not  already  a  factor  of  the  other  expression. 

For  a  factor  introduced  by  multiplication  into  one  expression 
will  not  be  common  to  both  of  them,  and  therefore  will  not  be 
introduced  into  their  H.  C.  F. 


122                                          ALGEBRA.  [Cn.  VI 

In  like  manner,  the  factor  removed  by  division  from  one 

expression  was  not  common  to  both  of  them,  and  therefore 
would  not  have  been  a  factor  of  their  H.  C.  F. 


Ex.  Find  the  H.  C.  F.  of  2^  +  5^-3  and 

2^  +  ^ 
We  have 

2x2  +  5x-3)2x3  +    x2 

-    3x 


_4a2-    2x 


The  next  step  would  introduce  fractional  coefficients.  To 
avoid  these,  we  divide  8  x  —  4  by  4,  since  4  is  not  a  factor  of 
2  x2  +  5  x  —  3)  and  take  2  x  —  1  as  the  divisor  of  the  second 
stage  : 


2X2- 


6x-3 
6a;.-3 

The  required  H.  C.  F.  is  2  x  -  1. 

37.  Before  proceeding  with  the  division,  remove  from  the 
given  expressions  any  monomial  factors  and  set  aside  their 
H.  C.  F.  as  a  factor  of  the  required  H.  C.  F. 

Ex.   Find  the  H.  C.  F.  of 


*  -  6  or3  +  6  x2  -  3  x  +  2), 
-  12  x2y 


We  set  aside  xy,  the  H.  C.  F.  of  2  xy2  and  3  x2y)  as  a  factor  of 
the  required  H.  C.  F.,  and  find  the  H.  C.  F.  of  the  remaining 
factors  by  division. 

The  first  of  these  expressions  cannot  be  divided  by  the 
second  without  introducing  fractional  coefficients.  To  avoid 


36-38]  H.C.F.   AND   L.C.M.   BY   DIVISION.  123 

these  we  multiply  the  first  by  2,  since  2  is  not  a  factor  of  the 
other  expression. 


7  x2  - 


+   4.x-    8 

-  35  x2 


2d  divisor,  5  x2  -  -    9  x  +   4 

To  avoid  fractional  coefficients  in  the  next  stage  of  the  work, 
we  multiply  the  last  divisor  by  5  : 

-    20(2x-7 


8a? 


X5)_    7X2+    27  x-    20 
1350;  -100 

-  28 

-  72 


3d  divisor,  x  —  l)5x*  —  9a;  +  4(5a;-  4 


-4a? 


To  avoid  fractional  coefficients,  we  multiplied  the  partial 
remainder  of  the  first  division  by  —  2,  divided  the  remainder 
of  the  first  division  by  5,  multiplied  the  partial  remainder  of 
the  second  division  by  5,  and  divided  the  remainder  of  the 
second  division  by  72. 

The  required  H.  C.  F.  is  xy(x  —  1). 

38.  If  the  divisor  and  dividend  at  any  stage  of  the  work  can 
be  factored  readily,  it  is  better  to  find  their  H.  C.  F.  by  factoring 
than  by  continuing  the  method  of  division. 

Ex.   Find  the  H.  C.  F.  of 

a-«  _  10  a3  +  35or>  -  50  x  +  24,  (1) 

and  tf-Tx2  +  llx-5.  (2) 


124  ALGEBRA.  [Cn.  VI 

We  have  : 


x4—  Tar'+ll^2—   5x 


9 


3 

The  remainder  x2  —  4  x  +  3,  =  (a;  —  1)  (a;  —  3),  is  readily  fac- 
tored. 

Dividing  Xs  —  1  Xs  +  11  x  —  5  by  x  —  1,  we  have 

^3  _  7  tf  +  n  x  ._  5  =  (g.  _  !)  (^  _  6  ^  +  5)  =  ^  _  ^2  ^  _  5^ 

The  H.  C.  F.  of  the  first  remainder  and  (2),  and  therefore 
the  required  H.  C.  F.,  is  x  —  1. 

Lowest  Common  Multiple  by  Means  of  H.  C.  F. 

39.   If  the  given   expressions  cannot   be  readily  factored, 
their  L.  C.  M.  can  be  obtained  by  first  finding  their  H.  C.  F. 

Ex.     Find  the  L.  C.  M.  of 

x*-2x2-2x?y  +  4:xy  +  x-2y  and  x*  -  2x2y  +  xy2  -  2f. 
The  H.  C.  F.  of  these  expressions  is  found  to  be  x  —  2  y. 

Consequently  the  other  factors  of  the  given  expressions  can 
be  found  by  dividing  each  of  them  by  their  H.  C.  F.     We  have 


From  the  definition  of  the  H.  C.  F.,  as  also  by  inspection, 
we  know  that  these  second  factors,  x2  —  2  x  +  1  and  x2  +  ?/2, 
have  no  common  factor,  and  therefore  that  the  L.  C.  M.  of  the 
given  expressions  must  contain  both  of  them  as  factors. 

Consequently  the  required  L.  C.  M.  is 


This  example  illustrates  the  following  principle  : 

The  L.  C.  M.   of  two   integral   algebraic    expressions    is    the 

product    of  their  H.  C.  F.    by    the    remaining  factors    of  the 

expressions. 


38-40]  H.'C.F.   AND  L.C.M.  BY  DIVISION.  125 

Relation  between  H.  C.  F.  and  L.  C.  M. 

40.  The  following  example  illustrates  an  important  relation 
between  the  H.  C.  F.  and  the  L.  C.  M.  of  two  integral  algebraic 
expressions. 

Ex.   TheH.C.F.  of 


and  a?8-  !  =  (»-  !)(«  +  !) 

is  (x  -  1). 

The  L.  C.  M.  of  the  same  expressions  is 


The  product  of  the  two  given  expressions  is 


In  general, 

The  product  of  two  integral  algebraic  expressions  is  equal  to  the 
product  of  their  H.  C.  F.  and  their  L.  C.  M. 

It  follows  from  this  principle  that  the  L.  C.  M.  of  two  integral 
algebraic  expressions  can  be  found  by  dividing  their  product  by 

their  H.  C.  F. 

EXERCISES  XIV. 

Find  the  H.  C.  F.  and  L.  C.  M.  of  the  following  expressions  : 

1.  a*  +  4  a  -  5,  x*-2x2  +  6x-5. 

2.  2  X3^r3x2-x-  12,  6  xs-17x2  +  2  x  +  15. 

3.  Xs  -  3  x  +  2,  y?  +  2  x2  -  x  -  2. 

4.  2  aj8  -  17  x2  +  19  x  -  4,  3  Xs  -  20  x2  -  10  x  +  27. 

5.  X3  _  5  tf  +  9  x  _  9?  X4  _  4  ,#  +  12  x  -  9. 

6.  x3  -  x2  -  9  cc  +  9,  cc4  -  4  x2  +  12  a?  -  9. 

7.  ^-3^  +  4,  aj8-2aj2-4oj  +  8. 

8.  ^_3x  +  2,  x4-6a2  +  8a-3. 

9.  2^  +  3^-2,  4^  +  16  aj2-  19  a?  +  5. 

10.  ^-3^  +  4,  3  z3-  18  ^  +  36  a  -24. 

11.  x3  —  (a  +  6  —  c)  ar>  +  (06  —  ac  —  be)  x  +  abc, 
Xs  —  (a  —  b  +  c)  ^  +  (ac  —  a6  —  6c)  a?  + 


126  ALGEBRA.  [Cn.  VI 


12.  or5  +  ar2-5#  +  3, 

13.  3  oj8  -  8  or2  -  36  a?  +  5,  9  x3  -  50  a?  +  27  a  -  10. 

14.  4«y-3ajy-4ajy  +  3,  5  or5?/3  +  8  afy8  +  a?y  -  14. 

15.  a*-3xy*-2y*,  2  y*  -  5  a^  -  xy2  +  6^. 

16.  a3  -  a2  -  5  a  +  2,  3  a3  -  a2  -  8  a  +  12. 

17.  xs  +  2  x2  +  2  x  +  1,  a8  -  4  x2  -  4  a?  -  5. 

18.  30.0?  —  25oor!  +  8a2a;  —  a3,  18  or5  -  24  a®2  +  15  a2a  -  3  a3. 

19.  2  a4  -3^  +  4  a2  -5z-  4,  2aJ4-a8  +  a-12. 

20.  403  —  So^  +  Sa;  —  3,  2#4-3a3  +  6a2-3a;  +  2. 

21.  4  a4  -  8  x3  -  3  x2  +  7  a  -  2,  3  x8  -  11  or2  +  2  a?  +  16. 

22.  36  a6  +  9  a3  -  27  a4  -  18  a5,  27  a562  -  9  as62  -  18  a462. 

23.  3^-10^  +  15^  +  8,   ^_2x4-6^  +  4^  +  13a 

24.  2<B8-3aj*_-8«-3,  2x4-  9^  +  13^  -23  x  -16. 
25. 


The  H.  C.  F.  and  L.  C.  M.  of  Three  or  More  Expressions. 

41.  To  find  the  H.  C.  F.  of  three  or  more  integral  algebraic 
expressions  find  the  H.  C.  F.  of  any  two  of  them,  next  the 
H.  C.  F.  of  that  H.  C.  F.  and  the  third  expression,  and  so  on. 

42.  To  find  the  L.  C.  M.  of  three  or  more  integral  algebraic 
expressions,  find  the  L.  C.  M.  of  any  two  of  them  ;  next,  the 
L.  C.  M.  of  a  third  and  the  L.  C.  M.  already  found,  and  so  on. 

EXERCISES   XV. 

Find  the  H.  C.  F.  and  the  L.  C.  M.  of  the  following  expres- 
sions : 


2.   or3- 

3. 

4. 

5.   2aj4-ar}  +  3orJ  +  #  +  4,  2x*  -3^-  2^  +  9x  -  12, 

4z4  -  16  or3  +  25  x2  -  23  x  +  4. 


40-43]    SOLUTION  OF  EQUATIONS   BY   FACTORING.     127 

SOLUTION  OF   EQUATIONS  BY  FACTORING. 
43.   The  roots  of  the  equation 

(*-l)(a!-2)  =  0  (1) 

are  evidently  1  and  2.  For  1  reduces  the  first  member  to 
0  x  (—  1),  =0;  and  2  reduces  the  first  member  to  1  x  0,  =0. 
Therefore  equation  (l).is  equivalent  to  the  equations 

x  —  1  =  0  and  x  —  2  =  0,  jointly. 

This  example  illustrates  the  following  method  of  solving  an 
equation  by  factoring : 

Transfer  all  terms  to  the  first  member.  Factor  this  first  mem- 
ber, and  equate  each  of  the  resulting  factors  to  zero.  Solve  the 
equations  thus  obtained. 

Ex.  1.    Solve  the  equation  x  (x  —  2)  (x  +  5)  =  0. 

Equating  factors  to  0,  x  =  0 ;  x  —  2  =  0,  whence  x  —  2 ; 
and  x  -f  5  =  0,  whence  x  =  —  5. 

The  roots  are  therefore  0,  2,  and  —  5. 

Ex.  2.    Solve  the  equation  x2  —  1  =  3. 

Transferring  3  to  first  member,  and  factoring,  we  have 
(a-2)(a;  +  2)=0. 

Equating  factors  to  0,  x  —  2  =  0,  whence  x  =  2 ; 
and  x  +  2  =  0,  whence  x  =  —  2. 

The  roots  are  therefore  -f-  2  and  —  2. 

The  statement  -f-  2  and  —  2  is  usually  written  ±  2,  read 
positive  and  negative  two. 

Ex.  3.    Solve  the  equation  x*  +  2  x  -  12  =  3. 
Transferring  3,  a*  +  2  x  -  15  =  0. 

Factoring,  (x  -f  5)  (x  -  3)  =  0. 

Equating  factors  to  0,  x  +  5  =  0,  whence  x  =  —  5 ; 
and  x  —  3  =  0,  whence  x  =  3. 

The  required  roots  are  therefore  —  5,  3, 


128  ALGEBRA.  [Cn.  VI 

EXERCISES   XVI. 

Solve  each  of  the  following  equations : 

1.  x(x-3)  =  0.  2.  x(x  +  5)  =  Q. 

3.  5x(x  +  7)  =  0.  4.  (a?  -  2) (a;  +  1)  =  0. 

5.  (3a?  +  2)(5a;--3)  =  0.  6.  a; (a;  +  2)  (3 a;  -  1)  =  0. 

7.  3  a;  (16  a2 -25)  =  0.  8.  (a?  -  1)  (9  or2  -  16)  =  0. 

9.  (^2-9)(4^-25)  =  0.    .   10.  (25x2-4)(x2 

11.  a2 -11  =  5.  12.  4^  —  15  =  1 

13.  23-9x2  =  -2.  14.  5«2-16  =  4 

15.  7^-46  =  5^  +  4.  16.  x2-x-2  = 

17.  x*  +  x  =  12.  18.  #2      3#-28 

19.  x2-x  =  30.  20. 

21.  So?2  — 13a?  — 10  =  0.  22. 

23.  15x2  +  14aj-8  =  0.  24. 

25.  (a?  -  5)  (a?  -  6)  =  30.  26.  (a- 12)  (a; +15)  = -180. 

27.  (a?  +  15)  0  +  4)  =  60.  28.  (aj  +  20)  (a?  -  5)  =  -  100. 

29.  If  24  is  added  to  the  square  of  a  number,  the  sum  will 
be  equal  to  eleven  times  the  number.     What  is  the  number  ? 

30.  If  40  is  added  to  the  square  of  a  number,  the  sum  will 
be  equal  to  thirteen  times  the  number.     What  is  the  number  ? 

31.  In  a  number  of  2  digits,  the  units'  digit  is  2  greater  than 
the  tens'  digit.     The  product  of  the  digits  is  equal  to  the  num- 
ber diminished  by  16.     What  is  the  number  ? 

32.  The  length  of  a  field  exceeds  its  breadth  by  3  rods.     If 
18  rods  were  added  to  its  length,  and  2  rods  were  taken  from 
its  breadth,  the  area  would  be  doubled.     What  are  the  dimen- 
sions of  the  field  ? 

33.  The  number  of  square  feet  in  the  area  of  a  square  floor, 
increased  by  20,  is  equal  to  nine  times  the  number  of  feet  in 
its  side.     What  is  the  length  of  a  side  of  the  room  ? 


CHAPTER  VII. 

FRACTIONS. 

1.  The  quotient  of  a  division  can  be  expressed  as  an  integer 
or  an  integral  expression  only  when  the  dividend  is  a  multiple 
of  the  divisor ;  as  a?b  •+•  ab  =  a ;  (ax*  -f-  2  bx)  -r-  x  =  ax  -f-  2  b. 

If  the  dividend  be  not  a  multiple  of  the  divisor,  the  quotient 
is  called  a  Fraction  ;  as  a  -=-  b ;    (ax2  +  2  bx)  -*-  a/-3. 

2.  The  notation  for  a  fraction  in  Algebra  is  the  same  as  in 
ordinary  Arithmetic. 

Thus,  (ax2  +  2  bx)  -v-  or5  is  written  aa;2  +  2  &» 

The  Solidus,  /,  is  frequently  used  instead  of  the  horizontal 
line  to  denote  a  fraction ;  as  (ax2  +  bx)/x?  for  - — •  t— ^. 

il/ 

3.  As  in  Arithmetic,  the  dividend  is  called  the  Numerator  of 
the  fraction,  the  divisor  the  Denominator,  and  the  two  are  called 
the  Terms  of  the  fraction. 

4.  An  integer  or  an  integral  expression  can  be  written  in  a 
fractional  form  with  a  denominator  1. 

E.g.,  1  =  \   a  +  b=±±*. 

It  is  important  to  notice  that  an  algebraic  fraction  may  be 
arithmetically  integral  for  certain  values  of  its  terms. 

E.g.,  when  a  =  4  and  6  =  2,  the  fraction  a/b  becomes  4/2  =  2. 

5.  By  the  definition  of  a  fraction,  a/b  is  a  number  which, 
multiplied  by  6,  becomes  a ;  that  is, 

(a/6).x6  =  a,  or  ?  x  b  =  a  (1) 

0 

129 


130  ALGEBRA.  [Cn.  VII 

6.   The  Sign  of  a  Fraction.  —  The  sign  of  a  fraction  is  written 
before  the  line  separating  its  numerator  from  its  denominator; 

"ff'-F       _ 

Since  a  fraction  is  a  quotient,  the  sign  of  a  fraction  is  deter- 
mined by  the  rule  of  signs  in  division. 

-}-«<__  .  a      —  a_     a     -\- a  _      a     —  a_      a 
+~&         &'    -b         b'    ~^b~~b'    ~+~b~~b' 


7.   From  the  rule  of  signs  we  derive : 

(i.)  If  the  signs  of  the  numerator  and  the  denominator  of  a 
fraction  be  reversed,  the  sign  of  the  fraction  is  unchanged. 

ZL!=.JL.    x     -x 

This  step  is  equivalent  to  multiplying  or  dividing  both  terms 
of  the  fraction  by  —  1. 

(ii.)  If  the  sign  of  either  tJie  numerator  or  the  denominator 
of  a  fraction  be  reversed,  the  sign  of  the  fraction  is  reversed; 
and  conversely. 

JP         7  _        —  7f      —  x  x         x  —  a x  —  a 

3"    '~3~'    x-l~~x-l'    ~b-x~x-  b 

(iii.)  If  the  signs  of  an  even  number  of  factors  in  the  numera- 
tor and  denominator,  either  or  both,  of  a  fraction  be  reversed,  the 
sign  of  the  fraction  is  unchanged;  but,  if  the  signs  of  an  odd 
number  of  factors  bV  reversed,  the  sign  of  the  fraction  is  reversed. 

x—a  x—a 


(a  _  b)  (b  _  C)  (c  -  a)         (a  -  b)  (b  -  c)  (a  -  c) 

x  —  a 

"  (b  _  a)  (b  -  c)  (a  -  c) 

a  —  x ^ 

~  (a-b)(b-c)(a-c) 


6-10]  FRACTIONS.  131 


Reduction  of  Fractions  to  Lowest  Terms. 

8.   A  fraction   is  said   to   be  in  its   lowest  terms  when   its 
numerator  and  denominator  have  no  common  integral  factor. 

Ea  2     X~l 

3'    rfTT 


9.    The  value  of  a  fraction  is  not  changed  if  both  numerator 
and  denominator  be  divided  by  the  same  number^  not  0. 

£  a  +  ab  _  (a  +  ab)  -r-  a  _  1  +  b 

a  +  ac      (a  -h  ac)  -T-  a     1  -f  c 

Let  the  value  of  -  be  denoted  by  v\  or  v  =  — 
b  b 

Multiplying  by  6,  vb  =  -  x  6  =  a. 

Dividing  by  n,     vb  -s-  n  =  a  ~  n,  or  v  (b  -j-  n)  =  a  •+•  n. 
Dividing  by  6  -5-  n,  v  =  a  -f-  n  -s-  (6  -r-  ?i), 


But  v  =  -- 

b 

Therefore 


b     b  -r-  n 
10.  Ex.  1.   Eeduce  ---  to  its  lowest  terms. 


The   factor   2a262  is   the   H.  C.  F.   of    the   numerator   and 
denominator.     We  therefore  have 


A  fraction  is  reduced  to  its  lowest  terms  by  dividing  its  nu- 
merator and  denominator  by  the  H.  C.  F.  of  its  terms. 

This  step  is  called  cancelling  common  factors,  and  can  usually 
be  done  mentally,  if  the  terms  of  the  fraction  are  first  resolved 
into  their  prime  factors. 


132  ALGEBRA.  [Cn.  VII 

Ex  2  a2  —  a?2  _  (a  +  a?)  (a  —  a?)  _  a  —  x 

(a  +  x)2  ~  (a  +  x)  (a  +  x)  ~  a  +  x 


Ex  3 


Changing  the  sign  of  the  first  factor  in  the  numerator  and 
the  sign  of  the  fraction,  we  have 

_   (2  y  -  x)  (x  +  y)   =  _ 


(2  ?/  -  x)  (2  y  +  x)          2y  + 


Ex.  4.   Keduce      -  -     to  its  iowest  terms< 

a?  —  or2  —  #  —  2 

We  find  x—  2  to  be  the  H.  C.  F.  of  numerator  and  denomi- 
nator by  Ch.  VI.,  Art.  33. 

Then 


(^  _  y»  _  x  _  2)  -i-  (x  -  2)        or*  +  a?  +  1 

EXERCISES  I. 

Eeduce  each  of  the  following  fractions  to  its  lowest  terms : 
1.    *.  2.    *%.  3. 


ac  "   ax2  '  Sa3^2  "   8ar3m2n6 


5'   5aW*  6       '"    '  "  '      7'     -  "-'          8' 


m  —  n  a2  +  ab  15  x  —  9 

'•   2wi-2w'  ^^6'  '   6 -10 a?' 

13.   ^±H2.  14.    2-^  "    5a2  + 


O  j.^.  (5  7-  -u^^.  0  =- 

x2  +  x  a^  —  4  a2  —  x2 

16.  J^il.  „.  «;  +  to          ia 

a3  —  63  wa2  —  nb2 

19.  ii^L.  20.;+^-;.    21. 

8  a6  —  2763  or  +  5  x  +  6 


22.      ,     '       n ^     23.   ^    „     I     , — s-^-     24 


3^_35_o5 

—     i    -  ~       —  _g    ur       ax   -|-  c>  x       ut> 

a*  -  aar  -  6*»  +  ab'2' 


10-12]  FRACTIONS.  133 


'  ' 


33^-14  x2n  +  3  xn  +  2 

,  **  +  *  30    a*-(b-c)2 

'•  22  '  22 


(g  +  fr)2-4c2  a2  +  a  +  b  - 

'  ' 


33.    »£Z»         if  *£  +  ?=*-- 

66  —  10  a  —  6  x2  —  if  - 


nc      -*-  —  ^ t  og        •*"          •"  if  —  •"y      \    y 

(1  +  ax)2  —  (ct  +  x)2  x4  —  x*y  —  x2y2  +  xy3 

37. 


5^_5a;  +  7 
39.          ^~®*  +  2       . 


*-] I  t  AO 

T3  ^      I      7*      I      1  A 

Jj  ds      ~~T~    dJ    "T"^    -LTt 


Reduction  of  Two  or  More  Fractions  to  a  Lowest  Common 
Denominator. 

11.   Two  or  more  fractions  are  said  to  have  a  common  de- 
nominator when  their  denominators  are  the  same. 

E.g.,  2  and  5;    -  and          *"»        . 

2       2 


The  Lowest  Common  Denominator  (L.  C.  D.)  of  two  or  more 
fractions  is  the  L.  C.  M.  of  their  denominators. 

E.g.,  the  L.  C.  D.  of  —  and  —  is  62c2. 
b2c         .be2 

12.    The  value  of  a  fraction  is  not  changed  if  both  numerator 
and  denominator  be  multiplied  by  the  same  number,  not  0. 

-Q  a  —  xa  —  xxa  +  x)       a2  —  x2 


a  -h  x      (a  +  a;)  x  (a  +  x)      (a  +  x) 


134  ALGEBRA.  [Cn.  VII 

Let  the  value  of  the  fraction  -  be  denoted  by  v,  or 


Multiplying  by  b,  vb  =  -  x  b  =  a. 

Multiplying  by  n,  vbn  =  an. 

Dividing  by  bn,  v  =  an  -+-  bn  =  — • 

But  v  =  2. 

b 

Therefore  -  =  ^L. 

b      bn 

13.   Ex.  1.   Reduce  —  and  —  to  equivalent  fractions  having 
b2c         be 

a  lowest  common  denominator. 
Their  required  L.  C.  D.  is  62c2. 

Multiplying  both  terms  of  -2-  by  b2c2-t-tfc.  =c,  we  have  — ; 

b2c  b'-cr 

and  both  terms  of  —  by  frV-f-te2,  =&,  we  have    — • 


Ex.  2.    Reduce   x,  =  -,  and  — ^—  to  equivalent  fractions 

having  a  lowest  common  denominator. 
The  required  L.  C.  D.  is  x  —  y. 

Multiplying  both  terms  of  -  by  x  —  y,  we  have  — 11^1 . 

1  x  —  y 

and  both  terms  of  — ^ —    by  1,  we  have,   — * — 

x  —  y  x  —  y 

Ex.  3.  Reduce 

1  1 


_  3  x  +  2'        (a?  -  1)  (x  -  2)' 


and  ^_     = 2 


to  equivalent  fractions  having  a  lowest  common  denominator. 
The  required  L.  C.  D.  is  (a;  -  1)  (a;  -  2)  (x  +  1). 


12-14]  FRACTIONS.  135 

Multiplying  both  terms  of  the  first  fraction  by 

we  have — ; 

(x  —  1)  (x  —  2)  (x  +  1) 

and  both  terms  of  the  second  fraction  by 

(aj  -  1)  (3 -  2)  (aj  +  l)-Ka?- !)(«  +  !),  =»-2, 
we  have  — — • 

14.   These  examples  illustrate  the  following  method : 
Take  the  L.  C.  M.  of  the  denominators  as  tlie  required  denomi- 
nator 

Divide  this  denominator  by  the  denominator  of  each  fraction; 
and  multiply  both  numerator  and  denominator  of  the  fraction  by 
the  quotient. 

EXERCISES  II. 

Reduce  the  following  fractions  to  equivalent  fractions  having 
a  lowest  common  denominator : 

1    x  4m   5n  5a*6   2a&* 

7  4  ~3~>  T  14  '  ~2T 

a2  l-i-4m  1  K       9  * 

4.  1  —  a?  -•        5.  m,  — : — —•          6. 

a •  -f- 1  m  —  4 

q  >7  9^/y^J_9/r 

7.   -4-,  1,  _-L_.    8.  ^^^,  5if£.    9. 


1  1  5 


1  3  •   2         3 

"  '  ' 


14.       1         ^L«   *4^  15. 


16  17 

'  ' 


__ 

?   1  -  x2 


136  ALGEBRA.  [Cn.  VII 

m  y          1  +  m    ig     ax—  b  a  —  bx  1 

y(x  —  y)'  m(y  —  x)J     my  ax  +  atf  bx  +  b2'  a~b'2 

3  5 


20.   r~f   -^— ,   =^h=-  21. 

22. 


Z7  a2 -a     a2-!  "  20-2'  a;2-2x-+l'   1-x2 

1          3  nm  m  —  n 


M    rm          W°    ffflV 

23. -i -,    - 


24. 
25. 


(a-c)(a-b)'   (b-a)(b-c)'   (c-a)(c-b) 

Equations. 
15.   Ex.  1.    Solve  the  equation  2  a;  +  -  =  9. 

Multiplying  by  4,          4  x  2x  +  4  x  |  =  4  x  9;  (1) 

or.  since  4  x  7  =  ic,  8  a;  +  x  =  36. 

4 

Uniting  terms,  9  x  =  36. 

Dividing  by  9,  x  =  4. 

The  step  represented  by  (1)  is  called  clearing  the  equation  of 
fractions,  and  should  be  performed  mentally. 

To  clear  of  fractions,  we  multiplied  by  4,  the  denominator 
of  the  fractional  term.  If  the  equation  contains  more  than 
one  fraction,  we  multiply  by  their  L.  C.  D. 

Ex.  2.    Solve  the  equation  -  —      t        =  3  —  x. 

The  L.  C.  D.  is  15. 

Multiplying  by  15,  3oj-5(2aj-l)  =  15  (3  -  x).      (1) 

Removing  parentheses,         3  x  —  10  x  +  5  =  45  —  15  x. 

Transferring  terms,  3x  —  10ic  +  15ic  =  45  —  5. 

Uniting  terms,  8  x  =  40. 

Dividing  by  8,  x  =  5. 


14-17]  FRACTIONS.  137 

16.  Observe  that  the  sign  of  a  fraction  affects  each  term  of 
the  numerator,  or  the  dividing  line  between  the  numerator  and 
the  denominator  has  the  same  effect  as  parentheses. 

Kg.,  _a-6  +  c==_(a_6  +  c)_f_d 

=  (—  a  +  6—  c)-r-d 
—  a  +  6  —  c 

d 

2  x  _  \ 

Thus,  in  Ex.  2,  Art.  15,  the  sign  —  before  the  fraction 


o 

changes  the  signs  of  both  terms  in  its  numerator,  and  not 
simply  the  sign  of  the  first  term,  when  the  denominator  is 
removed.  This  caution  should  be  kept  in  mind,  and  step  (1) 
omitted  in  clearing  of  fractions. 

17.   Ex.    Sol  ve  the  equation 

The  L.  C.  D.  is  24. 
Multiplying  by  24,     4 


Transferring  terms,        4  x  —  3  x  —  2  x  =  —  3. 
Uniting  terms,  —  x  =  —  3. 

Dividing  by  —  1,  x  =  3. 

EXERCISES  III. 

Solve  each  of  the  following  equations  : 


1. 

X 

x 

^2 

=  18. 

,  .-% 

=  5. 

3. 

10  + 

6  =  x. 

4. 

X 

2 

.x 
4 

=  15. 

2x     x 

3       2 

=  5. 

6. 

3x 
5  " 

H- 

X 

-2 

3-x 

o    a  —  4 

5- 

-  X        ~ 

3x-2     3x+2 

3 

2 

5 

4 

4 

5 

T  n 

X 

+  3 

\X         A 

x-l 

aj      o 

2 

4 

5 

1  o 

3 

/y»    

2     x     .. 

TO 

6«-5 

5x 

3 

5 

4 

4 

3 

14. 

3 

x 

a;  +  4 

1  *5 

5  a     a  -  10 

4 

6 

8           6 

138  ALGEBRA.  [Cn.  VII 


16  5  ft  —  1  _  g  1?    5—  a     a—  4  _  #—  3 

32  12          15        ~2<r* 

ft-6      ft-3      1     ft  +  1 


18. 


12        3        16 


19    5  —  x  _  x  —  3      a;  —  1  _  _7_ 
4  6  16        12* 

a;_4   _o!  —  3      1      ft-2 


3        12 


2i  - 

22 


18          20          24  3 

—  5     a;  —  3 


18          16          20 
Problems. 

18.  Pr.  In  a  number  of  two  digits,  the  units'  digit  is  two- 
thirds  of  the  tens'  digit.  If  the  digits  be  interchanged,  the 
resulting  number  will  be  18  less  than  the  given  number. 
What  is  the  number  ? 

Let  x  stand  for  the  tens'  digit  ; 

then  -f  x  stands  for  the  units'  digit. 

The  given  number  is  10  x  -f  f  x, 

and  the  resulting  number  is        10  x  -Jo?  +  x,  =  ^-x  +  x. 

The  problem  states, 
in  verbal  language  :  the  given  number  minus  the  resulting  num- 

ber is  18  ; 
in  algebraic  language  :  10  x  -f-  f  x  —  (-2^°-  x  +  x)  =  18. 

Removing  parentheses,         10  x  +  f  x  —  -\°-  x  —  x=  18. 

Clearing  of  fractions,          30  x  +  2  x  —  20  x  —  3  x  =  54. 

Uniting  terms,  9  a?  =  54. 

Dividing  by  9,  x  =  6, 

the  tens'  digit. 

Then  the  units'  digit  is  f  x,  =  4. 

Therefore  the  required  number  is  64. 


17-18]  FRACTIONS.  139 

EXERCISES    IV. 

1.  A  son  is  i  as  old  as  his  father,  and  in  18  years  he  will 
be  i  as  old.     How  old  is  each  ? 

2.  A  son  is  J  as  old  as  his  father,  and  6  years  ago  he  was 
i  as  old.     How  old  is  each  ? 

3.  A  boy  lost  4-  of  his  money,  and  afterward  found  12  cents. 
He  then  had  twice  as  much  as  at  first.     How  much  money  had 
he  at  first  ? 

4.  Two  men  invest  equal  amounts.     The  first  one  loses 
$  600,  and  the  second  one  gains   $  600.      The  first  then  has 
only  f  as  much  as  the  second.     How  much  did  each  invest  ? 

5.  Divide  65  into  3  parts,  so  that  the  second  shall  be  8 
greater  than  the  first,  and  the  third  f  of  the  sum  of  the  first 
and  second. 

6.  From  a  cask  full  of  water,  J  of  the  contents  is  drawn 
off.     If  10  gallons  are  then  poured  into  it,  it  will  contain  J  of 
its  original  contents.     What  is  the  capacity  of  the  cask  ? 

7.  The  sum  of  the  two  digits  of  a  number  is  12.     If  the 
digits  be  interchanged,  the  resulting  number  will  exceed  the 
original  one  by  three-fourths  of  the  original  number.     What  is 
the  number  ? 

8.  A  merchant  paid  30  cents  a  yard  for  a  piece  of  cloth. 
He  sold  one-half  for  35  cents  a  yard,  one-third  for  29  cents  a 
yard,  and  the  remainder  for  32  cents  a  yard,  gaining  $  18. lo 
by  the  transaction.     How  many  yards  did  he  buy  ? 

9.  A  woman  sells  ^  of  an  apple  more  than  one-half  of  her 
apples.     She  next  sells  \  of  an  apple  more  than  one-half  of 
the  apples  not  yet  sold,  and  then  has  6  apples  left.     How 
many  apples  had  she  at  first  ? 

10.  A  merchant  lost  -J-  of  his  capital,  and  then  \  of  what 
remained.     If  he  then  had  $12,000  capital,  how  much  had  he 
at  first  ? 

11.  Thirteen  coins,  dollars   and   quarter-dollars,  amount  to 
$  9.25.     How  many  coins  of  each  kind  are  there  ? 


140  ALGEBRA.  [Cn.  VII 

12.  A  box  contains  a  number  of  pencils,  of  which  J  are  red, 
-^  are  blue,  and  3  are  black.     How  many  pencils  are  red,  and 
how  many  are  blue  ? 

13.  The  deposits  in  a  bank  during  three  days  amounted  to 
$  77,700.     If  the  deposits  each  day  after  the  first  were  |  of 
those  of  the  preceding  day,  how  many  dollars  were  deposited 
each  day  ? 

14.  A  father  leaves  his  property  to  his  three  sons  as  follows  : 
to  the  first,  $  3000  less  than  1  of  his  property ;  to  the  second, 
$  2400  less  than  1  of  his  property ;  and  to  the  third,  $  1800 
less   than  i  of   his   property.     What   is   the   amount  of  his 
property  ? 

15.  A  father  divided  his  property  equally  among  his  sons. 
To  the  oldest  son  he  gave  $  300  and  TL  of  what  remained ;  to 
the  second  son  he  gave  $  600  and  -^  of  what  was  then  left ; 
to  the  third  son  he  gave  $  900  and  y1^  of  the  remainder ;  and 
so  on.     What  was  the  amount  of  his  property,  and  how  many 
sons  had  he  ? 

16.  The  height  of  the  first  platform  of  the  Eiffel  Tower  is 
8  meters  more  than  i  of  the  whole  height;  the  second  plat- 
form is  twice  as  high  as  the  first,  and  160  meters  less  than  the 
third ;  the  third  is  1  meter  greater  than  11  of  the  entire  height. 
What  is  the  height  of  the  tower,  and  of  each  platform  ? 

17.  Jupiter  has  1  more  moon  than  Uranus,  and  Uranns  half 
as  many  moons  as  Saturn ;  Mars  has  3  less  than  Jupiter,  and 
Neptune  half  as  many  as  Mars.     If  these  planets  together 
have  20  moons,  how  many  has  each  ? 

18.  A  leaves  a  certain  town  P,  travelling  at  the  rate  of 
21  miles  in  5  hours ;  B  leaves  the  same  town  3  hours  later  and 
travels  in  the  same  direction  at  the  rate  of  21  miles  in  4  hours. 
After  how  many  hours  will  B  overtake  A,  and  at  what  distance 
from  P? 

19.  A  train  runs  from  A  to  B  at  the  rate  of  30  miles  an 
hour ;  and  returning  runs  from  B  to  A  at  the  rate  of  28  miles 


18-20]  FRACTIONS.  141 

an  hour.  The  time  required  to  go  from  A  to  B  and  return  is 
15  hours,  including  30  minutes'  stop  at  B.  How  far  is  A 
from  JB? 

20.  A  servant  is  to  receive  $  170  and  a  dress  for  one  year's 
services.     At  the  end  of  7  months  she  leaves  her  place  and 
receives  $  95  and  the  dress.     What  is  the  value  of  the  dress  ? 

21.  A  cistern  has  three  pipes  which  can  empty  it  in  6,  8, 
and  10  hours  respectively.     After  all  three  pipes  have  been 
open  for  2  hours  they  have  discharged  94  gallons.     What  is 
the  capacity  of  the  cistern  ? 

22.  A  wall  can  be  built  by  20  workmen  in  11  days,  or  by 
30  other  workmen  in  7  days.     If  22  of  the  first  class  work 
together  with  21  of  the  second  class,  after  how  many  days  will 
the  work  be  completed  ? 

23.  At  6  o'clock  the  hands  of  a  clock  are  in  a  straight  line. 
At  what  time  between  7  and  8  o'clock  will  they  be  again  in  a 
straight  line  ?     At  what  time  between  9  and  10  o'clock  ? 

Addition  and  Subtraction  of  Fractions. 
19.   Add  -  to  -•     We  have 


This  proves  the  following  method  of  adding  two  or  more 
fractions  which  have  a  common  denominator: 

The  numerator  of  the  sum  is  the  sum  of  the  numerators,  and 
the  denominator  is  the  common  denominator. 

A  similar  method  is  applied  in  subtracting  fractions. 


Wn  --  X-l         i 

_£Z/tt/*«  •—  —   -  ^^  -  -  -  —  ~—~;  -  .  ^-77;    I 

x-l      x-l  x-l  x-l 

20.  If  the  fractions  to  be  added  or  subtracted  do  not  have  a 
common  denominator,  they  should  first  be  reduced  to  equiva- 
lent fractions  having  a  lowest  common  denominator. 


142  ALGEBRA.  [Cn.  VII 


Ex.1.    Simplify  --  +      • 
b2c     be2 

Wehave          a    I    d  -  ac 


Ex.2.    Simplify  2*-  5y  _3x-  6y  +  2s. 
5  4 

Reducing  to  L.  C.  D.,  we  have 
8a?  —  20y      15a;  — 


20  20 

8  a;  -20y-  (15  ^-30^  +  10  2) 
20 

_  8  x  -  20  y  -  15  x  +  30  y  -  10  s  =  —  7  a;  +  10  y  -  10  2 
20  20 

The  expressions  in  this  example  are  not  algebraic  fractions. 

The  beginner  should  be  careful  in  subtracting  a  fraction  to 
change  the  sign  of  each  term  of  the  numerator,  and  not  that  of 
the  first  term  only. 

In  like  manner  we  may  change  the  sign  of  each  term  of  the 
numerator  (or  denominator),  if  we  change  the  sign  of  the 
fraction.  Thus,  in  the  result  of  Ex.  2,  we  have 


20  20 

Ex.3.   Simplify^- ?-+**-. 


The  L.  C.  D.  is  a?  -  1. 

Therefore,  -i-  — 2-+    3*  - =    + 

OJ  —  1      a;  +  1      y?  —  1      ar  —  1 


20]  FRACTIONS.  143 

EXERCISES  V. 

Simplify  the  following  expressions : 

i      ?_,.*  2     ?  +  — •  3     A    __L.  4     1      I. 

'    fc  +  a  '    8^16  '    4n      6n  '    a&^ac 


|2  1 j   9  C          I  T ^   .  ^ I      _      ^ 

xy     xz     yz  2  a     4  a     6  a  ab2     orb 

8.    3a?  +  5  +  3^~lt  g^ 


32  64 

3  a?  —  2     #  +  7  a  —  3     a  —  5     4  —  a 

10.   — = : h*.  U.    — FJ— 


J  6            8 

a_l     x-2     aB  +  7           __  3-2a  3a-2     6a  + 

12>   "2~     ~3~     ~6~                   ~3~  ~5~^         10" 

5-3a;    5a;-4    25-19  a?            x-2  2x  —  5     4-3^ 

14>   ~T~     ~10~       T5~  '           ~3^~"  ^^ 


~5~  "HT" 

x  —  y  —  z     5y  —  3z  —  x 


47  14 

18.    X  ~     ® 


3a-46     2a-6-c     15a-4c     a  — 

^~  ~~ 


12  21 

20. — -         21.    -^T  +  2~^-T'  22-    ^Ill_^^|. 

23.    £ll_«-2  24.    W±^_!^.          25.    _J» JL_ 

ic+2                m—n     m-\-n  aw4-l     aw— . 


27  | 

' 


a2  —  1     a  +  1  a2  —  4  ?/2     ac  +  2  cy 

3a-l      l-3a     3a-16 


a2-9       a+3 


144  ALGEBRA.  [Cn.  VII 

3  a          a         2  ax  x—  1         \—x         x—5 

\ 


a  +  x     a-x     a2-x2  6#+24     or'-lG     3z-12 

2m-3 ,3,2  2 ,    x-6  1 

32.      ~T~  • ~~r~  "  OO.       —  —I • 

1  —  4  m2     1  —  2  m     ra  & 


34. 


9  a2  -  25  62     6  ad  +  10  6d     6  ad  -  10  6d 

2 3  |  4 

4^  2/K 

36.    — 


35. 


^-5  ax      2  a2 


37.    -^-r-     3       ,    •  38.    — L-—  + 

a  — I  a3  —  1  x  +  1 

a  —  2n  a  —  n  1 

a3  +  ^3  »2n  —  art2  +  n3     an  -f-  w2 

1  3  nra  m  —  n 

n  —  m  n3  —  m3     m2  -f-  mn  -\-  n2 


«.  -4=  -^  -+- 


oj_8     x-7  '  x  +  S 

21.  Frequently  the  denominators  are  multinomials  in  the 
same  letter  of  arrangement,  but  not  arranged  to  the  same  order 
of  powers. 

TT,           c<-      TJ?        x           2x          2x 
Ex.  1.    Simplify  — -. 1 -• 

It  is  better  first  to  change  the  second  fraction  so  that  the 
denominators  shall  be  arranged  in  the  same  order.  We  then 
have,  by  Art.  7  (ii.), 

x  2x          2x  t 

x - l      tf -  I     x  +  1 

The  L.  C.  D.  is  x2  -  1. 


20-21]  FRACTIONS.  145 

Therefore, 

x  2x          2x    =a?(o;  +  l)—  2a;  —  2a?(a?  —  1) 

~ 


or* 


x2  —  1  or  —  1  05  +  1 

As  in  this  example,  the  result  of  the  addition  should  be 
reduced  to  its  lowest  terms. 

Ex.  2.    Simplify 


(a— b)(a— c)      (b—a)(b—c)      (c—d)(c—b) 

Changing  the  fractions  into  equivalent  fractions,  whose 
denominators,  taken  in  pairs,  have  one  common  factor,  we 
have 


(a  —  b)  (a  —  c)      (a  —  b)(b  —  c)      (a  —  c)  (b  —  c) 

b  —  c  a  —  c 

(a  -  b)  (a  -  c)  (b-c)~  (a-  b)  (b  -  c)  (a  -  c) 

a  —  b  _  _  b  —  c  —  a  -}-  c  -\-  a  —  &_ 

(a  —  c)  (b  —  c)  (a  —  b)  ~  (a  —  b)  (a  —  c)  (b  —  c)  ~~ 


EXERCISES   VI. 

Simplify  the  following  expressions  : 
36  a 


3. 


5. 


7. 


X2- 

4     2-oj 

3 

—  a     a2  —  3  a 

b 

a 

4 

a_l_4         ^_2 

a2- 

ab      b2  —  ab 

5 

o;  —  10      6  -  3  o; 

3 

7 

4-20  x 

m            2mn 

2m 

2x—l  '  2#+l 

l-4ar> 

m  —  ?i      w2  —  m2 

m  +-w 

a  — 

I/     .  -i 
/  £J  -4-    X 
1                            1 

a2  +  1\ 

1                    1 

a  + 

1      1^1  -  a  ' 

a2  —  lj 

iC: 

B-3oj-f2     1- 

it-2 

1 

1 

,          1 

+  a;2  +  1      a  -  1  -  a2  'a?  +  1  +  or2 


10-    ;  i r^  + 


ALGEBRA.  [Cn.  VU 

1  1 


11   fl+± + *±£ i        c  +  <* 

(b  -  c)  (c  -  a)      (c  -  a)  (a  -  6)      (a  -  b)  (b  - 
12.  ab + *° +  ca 


(6-c)(c-a)      (c-a)(a-b)      (a-b)(b- 
13.  -^       ,+  1  '  1 


a(a-6)(a  — c)      b(b-a)(b-c)      c(c-a)(c  — 6) 


14.    !  \ + ^ -f H 

(a  —  6)  (a  —  c)      (b  —  a)  (b  —  c)      (c  —  a)  (c  — 

15.  ^-     -+  *  '  * 


(a  —  b)(a  —  c)      (b  —  a)  (b  —  c)      (c  —  a)  (c 

16    be ac 

a(a2  -  b2)  (a2  -  c2)      b  (b2  -  a2)  (62  -  c2)     c( 


17>          a2-6c         ,         62  +  ac         ,         c2  +  ab 


Reduction  of  Mixed  Expressions  to  Improper  Fractions. 
22.   A  Proper  Fraction  is  one  whose  numerator  is  of  lower 
degree  than  its  denominator  in  a  common  letter  of  arrange- 
ment. 


E.g., 


oj  +  +    ^- 

An  Improper  Fraction  is  one  whose  numerator  is  of  the  same 
or  of  a  higher  degree  than  its  denominator  in  a  common  letter 
of  arrangement. 

Wn 


If  both  integral  and  fractional  terms  occur  in  an  expression, 
it  is  sometimes  called  a  Mixed  Expression. 

23.   Ex.  1.    Eeduce  a-f--  to   an   improper  fraction.     First 
c 

reducing  a  to  the  form  of  a  fraction  with  denominator  c,  we 
have 

.b_acb_ac-)-b 
c      c       c          c 


21-24]  FRACTIONS.  147 

This  example  illustrates  the  following  method : 
Multiply  the  integral  part  by  tJie  denominator  of  the  fractional 
part.     To  this  product  add  algebraically  the  numerator  of  the 
fractional  part)  and  write  the  sum  as  the  required  numerator. 

Ex.  2.    Simplify  1  —  x  +  x2  — 


Wehave  1  -  x 


x 


1  +  a;  1  +  a; 


EXERCISES  VII. 

Simplify  the  following  expressions  : 


1.   2a-§.                      2.    7+i- 

3.    m+i. 

3                                   a 

m 

4a         X    1     1                                 *»       1      I 

6        3  <7     1 

JO                                                                      t)u          J_ 

3 

7    °m      3m~5^-      B    a         ^ 

oil      8a 

4                           a  +  6 

a  -  ft)2 

0              „ 

O  —  X 

12.   a  +  4      9ic  +  20                   13    5a: 

42  —  a; 

14      ^y.  -U  7)                                                       IB      «        7 

a-7 

4  a/j 

i  j  —  —   . 

a—b  a—b 

16.    I -fa-— — \  17.   a2  +  a.T 


1VW  !         v      „         ,  , 

+  ay  a  —  x 

Reduction  of  Improper  Fractions  to  Mixed  Expressions. 

24.  Ex.  l.   Reduce  -  '  x  '  °  to  a  mixed  expression. 

x  +  1 

We  have  2  a?2  +    x  +  5  I     a;  +  1 


-    aj-1 


148  ALGEBRA.  [Cn.  VII 

But  by  Ch.  III.,  Art.  47,  we  have 

(2x*  -f-  x  +  5)  -r-  (x  +  1) •  =  2x  -  1  +  6  -s-  (a;  +  1), 

or  =  2  #  —  1  H -• 

This  example  illustrates  the  following  method : 

Divide  the  numerator  by  the  denominator,  until  the  remainder 
is  of  lower  degree  than  the  divisor. 

Write  the  remainder -as  the  numerator  of  a  fraction  whose 
denominator  is  the  divisor. 

Add  this  fraction  to  the  integral  part  of  the  quotient. 

Ex.  2.   Eeduce  — "*"  ^  ~ — x^~     to  a  mixed  expression. 

x2  +  2  x  —  1 

We  have 


x  -I 


-x  +  2 
Therefore,    ^  +  ^~  **  +  3  =  x-  1  + 


EXERCISES  VIII. 

Reduce  each  of  the  following  fractions  to  equivalent  frac- 
tional expressions,  containing  only  proper  fractions  : 

_1  10a2-3a 


6a8  — 


3  a  x  —  y  a  —  b 


_  ^  a;-5      .    21  xz  +  20  x  - 

a-1  oj  +  1  3x  +  2 

ms-ns-l  1;i    a;3  -3  o^  +  2  a;  -3 


24-26]  FRACTIONS.  149 

m3—  mn2—  m2n+?i3+l  _0    5xz  —  3  #  —  14 


_. 


.. 

m-n  x?-2 

1a?  +  9  (_    ar3  +  a;2  -  2 


Multiplication  of  Fractions. 

25.  Multiply  -  by  -.     Let  the  value  of  -  be  denoted  by  v, 

and  that  of  -  by  w  ;  or 
a 

(I  i  C 

v  =  -,  and  w  =  — 
b  d 

Multiplying  the  first  equation  by  b,  and  the  second  by  d, 
we  have  vb  =  ^  and  wd  =  ^ 

Multiplying  together  corresponding  members  of  these  equa- 
tions, we  have 

vb  x  wd  =  ac, 

or  vw  x  bd  =  ac. 

Dividing  by  bd,  vw  =  ac-t-bd=  —  • 

bd 

But  vw  =  -x-- 

b     d 

Therefore  ^x«  =  ^. 

b     d     bd 

This  proves  the  following  method  of  multiplying  fractions  : 

The  numerator  of  the  product  is  the  product  of  the  numerators; 
and  the  denominator  of  the  product  is  the  product  of  the  denomi- 
nators. 

26.  Ex.  1.   Find  the  product 

22  xY     25  a'2b 

The  factor  5  is  common  to  the  numerator  of  the  first  fraction 
and  the  denominator  of  the  second.  Since  to  cancel  a  com- 
mon factor  before  multiplication  is  equivalent  to  cancelling  it 
after  the  multiplication,  we  should  first  cancel  5.  For  a  similar 


150  ALGEBRA.  [Cn.  VII 

reason  we  should  cancel  the  factors,  2,  a1,  b,  x,  and  y-  before  the 
multiplication.     We  then  have 


3  ab       1      21  ab 

X    T~ 


11  xy3     5      55  xy* 

In  general,  if  the  numerator  of  one  fraction  and  the  denomi- 
nator of  another  have  common  factors,  such  factors  should  be 
cancelled  before  the  multiplications  are  performed. 

8a2 


Ex.  2.    Find  the  product 

a2  —  b2         4  a 

Cancelling  the  common  factors,  4  a  and  a  -\-b,  we  have 
2q        a +  6     2a(a  +  b} 

' /\ 


a  —  b 

27.   Ex.   Find  the  product  X9  ~  y9  x  (x  + 

ar  +  r 


x  — 
We  have 


Observe  that  a  fraction  is  multiplied  by  an  integer,  by  multi- 
plying its  numerator  by  the  integer. 

28.   If  one  of  the  factors  is  a  mixed  expression,  it  should 
first  be  reduced  to  an  improper  fraction. 

Ex.   Find  the  product  fj Y— — -  Y 

x  —  1          1 


WTe  have 

\      xj\x*  —  iy       x 

1 


i     "1 

EXERCISES   IX. 

Multiply : 
1.  -  X  3.  2.   —  X  5.  3. 


ay 


26-28] 
7. 


FRACTIONS. 


_    15  oW      14  xy2       _ 

I         NX    ••'     ,  Q  '      '  "     t7      y 

25«26  21a26 


151 


4«2     156     26c_ 
°'  565  x  -fj  x  3^ 


u.    *    xs»ax 

2  62c2       ax3 


12. 


14. 


16. 


18. 


x(3a-26).       13.  -J^Ux 


a62  —  ^>3     a3  —  aft2 


a2  4- 


262 


a2-62       2  a 
15.    _  ^      ^,     X  ?- 


5  a&  (a 


17. 


19.  ^Ix 


X 


20.        «(«+&)      x      b(a-b) 


6  ax—  15  5ic      8  ax  + 


4a2-2562 


22. 
23. 
24. 
25. 


x  —  y 


x  + 


x2  —  (a  +  c)  x  +  ac      x2  —  b2 
a2-(b-c}2         (x 


(a  _  6)2  _ 


26.  -^    X 

x2-y2 


27. 
28. 
29. 


-  4 


-  9 


x2-8x  + 15     ^-8;c  +  12 
a?  —  y  -f  «     ic2  +  2  #?/  -f-  ?/2  —  z 


12  a2 


4  a;2  —  9  y2    , 

22  a2  -  10  ab  *   6  ax  -9  ay   ^10  60;  +  15  6y" 


152 


ALGEBRA. 


[Cn.  VII 


' 
«, 

«5-L. 


a  —  6       x2  +  x  -  12      ff2  -  3  x  -  10 
X 


x  -  20 


x  -  6  x2  -  4 


(m  -f-  w)3         12  m  —  n  x  -\-  y 

32.     (o?_  OJ4-1M-.4--4-1  )•       33.    (- 


34. 


a 
36. 


35- 


-{- 


37. 


38.      a2  - 


-2/2 


m2  —  (6  —  a)2      (m  —  a)2  —  62     am  —  afr  +  a2 

' 


40.      1  + 


x 


y-zj     (x 
2-4c2 


Reciprocal  Fractions. 

29.  The  Reciprocal  of  a  fraction  is  a  fraction  whose  numera- 
tor is  the  denominator,  and  whose  denominator  is  the  numera- 
tor, of  the  given  fraction. 

E.g.)  the  reciprocal  of  -  is  — 
b       a 

30.  The  product  of  a  fraction  and  its  reciprocal  is  1. 


For 


b     a     ba 


28-32]  FRACTIONS.  153 


Division  of  Fractions. 

31.   Divide  -  by  -•     Let  the  value  of  -  be  denoted  by  i>, 
b        d  b 

and  that  of  -  by  w  ;  or, 

a        -i          c 

v  =  -,  and  w  =  — 
b  d 

Multiplying  the  first  fraction  by  b,  and  the  second  by  d}  we 

have 

v  b  =  a,  and  wd  =  c. 

Dividing  the  members  of  the  first  equation  by  the  corre- 
sponding members  of  the  second,  we  have 

vb      a 

vb  -r-  wd  =  a  -j-  c,  or  —  =  — 
wd     c 

Multiplying  byf,      5xf-fx£, 

v      ad 

or  -  =  —  • 

w      be 

But  -  =  -  +  r 

w      &      d 

Therefore  «  +  «=^  =  «xi 

o      a      be      b      c 

This  proves  the  following  method  of  dividing  one  fraction 
by  another  : 

Multiply  the  dividend  by  the  reciprocal  of  the  divisor. 
32    Ex  1    4(a2-a6)  .      6a     =4a(o-6)      (a-6)(a  +  6) 
(a  +  &)2     '  a2-62        (a  +  &)2  -6a 

=  2  (a  -  b)2 
3  (a  +  b)  ' 


a?  4-  2/2     »  -  y     ^2  +  2/2 

Observe  that  a  fraction  is  divided  by  an  integer  by  dividing 
its  numerator  by  the  integer. 


154 


ALGEBRA. 


[Cn.  VII 


EXERCISES  X. 
Simplify  the  following  expressions  : 


2       ^4.      3    6 

a  6 


.   2 


2* 


5  by 


^ 
15  / 


' 


7 

' 


' 


21 

12  a 


' 


2  a3  - 


a  - 


a  +  26         2a  +  4&  7(or3-l)    "    (1  -  a?) 

a2  -  (&  -  c)2  .  a  -  b  +  c  a8-!   . 

'•  : 


1  +  n  -  n3  -  n4  .  n2  -  1 

:  ' 


19. 
20. 
21. 
22. 
23. 


a;  — 


a?2  H-  ?/2 


or3  —  s/3          '  x?  —  y  (2x  —  y) 


25 


(q?  +  m)2  -  (.y  +  7i)2  .  (a;  -  y)2  -  (n  -  m 
(x  +  y)2  -  (m  +  n)2  '  (a?  -  m)2  -  (n  -  2/) 


32-34]  FRACTIONS.  155 


Complex  Fractions. 

33.  A  Complex  Fraction  is  a  fraction  whose  numerator  and 
denominator,  either  or  both,  are  fractions. 

2  a+x     1      1 

3  a  —  x  x 

E.G.*  ~TJ      -  ;  -  ?      -  T* 

4  a  +  y     ^  __1 

5  a  —  y  x 

Observe  that  the  line  which  separates  the  terms  of  the  com- 
plex fraction  is  drawn  heavier  than  the  lines  which  separate 
the  terms  of  the  fractions  in  its  numerator  and  denominator. 


34.   Ex.  1.   Simplify  -^ — 
1  —  x 

Multiplying  both  numerator  and  denominator  by  a?,  we  obtain 


x.  —  x)x.  —  x)          x 
To  reduce  a  complex  fraction  to  a  simple  fraction  : 

Multiply  both  its  terms  by  the  L.  C.  D.  of  the  fractions  in  the 
numerator  and  denominator. 


Ex.  2. 


x  +  — L_    *+   *       .*1=S 

1+?L±J:  _J_  4 

^3-x  3-x 

3  4 


3a?  +  3      x  +  1 
4 

Observe  that  in  this  reduction  the  work  proceeds  from  below 
upward. 


156  ALGEBRA.  [Cn.  VII 

EXERCISES  XI. 

Simplify  the  following  expressions  : 

!  ax  x      x  +  l 

2.    ?Ltf .  3>    »_1 » 


a  —  x 


4.  -     —  .  5.   a  +  -.-  6.  a 


1+a;  a 


_a_  +  1  3. 1_  ^_&_2 

—  1  1     _1_  /*  7>3          ,,3 


8.    , 1-5.  9. 


1  —  a  aj  + 1  a 

1  a  +  x        2x 


la  x_ 


1+ .  a?  a:  +  2  a2  + 


4— a;  1  —  2  a;  x  —  a 

a  -\-x     x  —  y  1  1 


13.        "...  .    ,.g     •      14.    1  + ^T-T     15.    1- i 


1- 


1—  a;  1—  # 

a  +  x     a  —  x  a  +  1      a  —  1 

,a  —  a;     a  +  x  a  —  I      a-f-1 

16.    -  •       17.    -  -  -  •  18.    -  •  —  -  -  -• 

n             n                          4  ax  a  +  1      a  —  1 


n  —  x     n  -\-x  a2  —  x2  a  —  1      a-f-1 

x  x  x*  1 

x-2 


+-+! 

OB  '  ar3 
a     n  —  x  ,       ax  f       1\Y        IV 


^         a         n*-  nx  V       gy  V       g. 

a          n-a;         '  '    /        1V/        I 


-  x         a  V       PJ  V       JP. 


34]  FRACTIONS.  157 

EXERCISES  XII. 
MISCELLANEOUS    EXAMPLES. 

Simplify  the  following  expressions: 


3 

a 


4.  ^+«r^W»±»-+2=i\ 

\c  +  d     c  —  dj     \c  —  d     c  +  dj 

11                          a  &  2 

5.    a  +  & r T-  6. +  — 


7.   m  — 


-     u-\--  -L  -r-     AT-      -TT 

6  a  6  a     a     6 

1 


1—  ra+w2— 


1  +  m 


,   ^  +  a)^_.)_(^  +  (,)(_±__«). 


10.  i 11. 


1_2      J^a  +  z 

«  I         o 


03  —  1  '  x  +  1  a2     ax     x2 

n-l      n  +  lY   A      w       1 


T  O  

X<S.      i    —  T    \  s^\    --  -- 

4      47i 


ab  +1 


13.  a  +  ^      a2  +  n2  +  2«n       14> 


a  +  n      a2  -  n2 


a+i     &+i 


a+ 


T  v  ~ VH 9  ^  9      .  o 

6             c             a               x      x2  +  a2     a;      x2  +  a2 
15.    -r  X ^X T-       16.    3 • h 


iJ         ^     HI"  1       a  +  s"1"!       a-a? 

i —      ^~TT      ^*i —  9        9  "    ••>  i     -j 

a  6  c  aa-  +  araa"  +  or 


158 


ALGEBRA. 


[Cn.  VII 


17.       _  +  A 


18. 


19. 


x  1  + 


2  be 


In  each  of  the  following  expressions  make  the  indicated 
substitution,  and  simplify  the  result : 


m  —  tt 


22.  In  . 

^m  —  b 

23.  In  1  + 


-,  let  a  +  &  +  c  =  2  s. 


24.    In  ^l-^-1-         let  «  = 
»  a       m          a 


Verify  each  of  the  following  identities 


= 


37. 


-z),  when 


CHAPTER   VIII. 

FRACTIONAL   EQUATIONS   IN   ONE  UNKNOWN 
NUMBER. 

1.  A  Fractional  Equation  is  an  equation  whose  members, 
either  or  both,  are  fractional  expressions  in  the  unknown 
number  or  numbers. 

E.g, 


x  +  2     a  +  1  a?  +  1 

3  2 

2.  Ex.  1.  Solve  the  equation  -  -  =  --  -• 

Multiplying  by  (a:  +  !)(«  +  2),     3  (a?  +  1)  =  2  (a?  +  2). 
Transferring  terms,  3  #  —  2  #  =  4  —  3. 

Uniting  terms,  a?  =  1. 


In  clearing  this  equation  of  fractions,  we  multiplied  by  an 
expression,  (x  -f-  1)  (x  +  2),  which  contains  the  unknown  num- 
ber. In  such  a  case  a  root  may  be  introduced.  But  it  is 
proved  in  School  Algebra,  Ch.  X.,  that,  if  a  root  is  introduced 
in  clearing  of  fractions,  it  must  be  a  root  of  one  of  the  factors 
of  the  L.  C.  D.  equated  to  0.  Since  1  is  not  a  root  of 

x  +  1  =  0,  or  of  x  +  2  =  0, 
it  is  a  root  of  the  given  equation. 

2  x  -\- 19        17               3 
Ex.  2.    Solve  the  equation  ^ = • 

The  L.C.D.  is  5  (x2  - 1),  =  5  (x  -  1)  (x  +  1). 

159 


160  ALGEBRA.  [Cn.  VIII 

Multiplying  by  5  (a;2  -  1),  2  x  +  19  -  85  =  -  15  x  -  15. 
Transferring  terms,  2  x  +  15  x  =  —  15  —  19  +  85. 

Uniting  terms,  17  x  =  51. 

Dividing  by  17,  a;  =  3. 

Since  3  is  not  a  root  of  x  —  1  =  0,  or  of  x  •+•  1  =  0,  it  is  a 
root  of  the  given  equation. 

Ex.  3.    Solve  the  equation  6x  +  l  -  2a7"1  =  3x~l. 

4          3x-2          2 

When  the  denominators  of  some  of  the  fractions  do  not  con- 
tain the  unknown  number,  it  is  usually  better  first  to  unite 
these  fractions. 

Transferring3^*,  «£L±1. 

Uniting  first  two  fractions, 

Multiplying  by  4  (3  x  -  2),  9  x  -  6  -  8  x  +  4  =  0. 

Transferring  and  uniting  terms,  x  =  2. 

Since  2  is  not  a  root  of  3  x  —  2  =  0,  it  is  a  root  of  the  gfren 
equation. 

Ex.  4.    If  both  members  of  the  equation 


be  multiplied  by  x2  —  1,  we  obtain  the  integral  equation 

-  2  x2  -  x(x  +  1)  =  -  x(x  -  1)  -  3  (x2  -  1), 
3r  (aj  +  l)(«-3)  =  0.  (2) 

Now  observe  that  it  was  not  necessary  to  multiply  by  x2  —  1, 
=  (x  +  !)(#  —  1),  to  clear  the  given  equation  of  fractions.  For, 
if  the  terms  in  the  second  member  be  transferred  to  the  first 
member,  we  have 


_       _ 
or,  uniting  terms,  —  —  -  =  0, 

XT  —  1 


2]  FRACTIONAL  EQUATIONS.  161 

y.     g 

or,  cancelling  x  +  1,  — —  =  0. 

Clearing  the  last  equation  of  fractions,  we  have 

3-3  =  0;  (3) 

whence  x  =  3. 

The  root  3  of  the  derived  equation  (3)  is  found,  by  substi- 
tution, to  be  a  root  of  the  given  equation.  Had  we  solved 
equation  (2),  we  should  have  obtained  the  additional  root  —  1, 
which  is  not  a  root  of  the  given  equation. 

This  root  was  introduced  by  multiplying  both  members  of  the 
given  equation  by  the  unnecessary  factor  x  -f- 1,  and  is  a  root  of 
the  equation  obtained  by  equating  this  factor  to  0. 

EXERCISES  I. 

Solve  each  of  the  following  equations : 

-2     3 


JM 

4. 
6. 
8. 
10. 
19 

x-3~ 
5 

7 

»  — 

oj  +  2 

5 

5. 

7 
9 
11 
1.^ 

x-12 

7 

24-  a;     - 
3 

x  +  17 

x  +  7 
x-l 

4 

2o?  +  3 

3x  +  5 

2 
0±1 

rr 

/ 

6 

6x  +  2 
x         1 

14 

=  5.  3. 


.34 

7 


0  —  8     x  —  < 
11  9 


3^-2 


5          6^  +  3         15 
;-l      l-2x 


6          l  +  2a? 
13.  -^fL.   _9_  =6. 


3  (  a?-5  =  2>  15    a?-9  (  ^- 


16    -*-  +  -^. 

'  x  +  2     x-2  a,*2  -  4               "  0  +  3. . '  ~  0 .  -  3  x2- 

R       ,   a»  J.1            a*1                                ^                7  19 

__              O                ^/^-J.                 »(/  _rt             «J.«  JLA 


x +. 


162  ALGEBRA.  [Cii.  VIII 

x         x-l2x2-5x-2l 

' 


1  .  x-4:2x*-7x-29 


23. 
24. 
25. 
26. 

28. 


Problems. 

3.  Pr.  1.  A  number  of  men  received  $120,  to  be  divided 
equally.  If  their  number  had  been  4  less,  each  one  would 
have  received  three  times  as  much.  How  many  men  were 
there  ? 

Let  x  stand  for  the  number  of  men.     Then  each  man  re- 

120 

ceived  —  —  dollars.     If  their  number  had  been  4  less,  each  one 
x  19() 

would  have  received  -  —  dollars, 
a  — 4 

The  problem  states, 

in  verbal  language:  the  number  of  dollars  each  would  have 
received,  if  there  had  been  four  less,  is  equal  to  three  times 
the  number  of  dollars  each  received. 

in  algebraic  language :    =  3  x • 

x  —  4  x 

Whence,  x  =  6. 

Therefore  there  were  six  men. 


2-3]  FRACTIONAL   EQUATIONS.  163 

Pr.  2.  A  can  do  a  piece  of  work  in  9  days,  B  in  6  days ;  and 
A,  B,  and  C  together  in  3  days.  In  how  many  days  can  0  do 
the  work  ? 

Let  x  stand  for  the  number  of  days  it  takes  C  to  do  the  work. 

Then,  in  one  day, 

A  does  -  of  the  work;  B  does  -;  and  C  does  -• 
9  6  x 

In  3  days, 

3  33 

A  does  -  of  the  work;  B  does  -;  and  C  does  — 
9  6  # 

Therefore,  in  3  days,  A,  B,  and  C  together  do 

?  +  ?  +  §  of  the  work. 
9     6      # 

The  problem  states, 

in  verbal  language :  the  work  done  by  A,  B,  and  C  together  in  3 
days  is  equal  to  all  the  work,  or  1 ; 

000 

in  algebraic  language :      -  +  -  +  -  =  1. 
y     o     x 

Whence,  x  =  18. 

Therefore  C  can  do  the  work  in  18  days. 

Pr.  3.  A  cistern  has  3  taps.  By  the  first  it  can  be  emptied 
in  80  minutes,  by  the  second  in  200  minutes,  and  by  the  third 
in  5  hours.  After  how  many  hours  will  the  cistern  be  emptied, 
if  all  the  taps  are  opened  ? 

Let  x  stand  for  the  number  of  minutes  it  takes  the  three  taps 
together  to  empty  the  cistern. 

Then,  in  1  minute,  the  three  together  will  empty  -  of  the 
cistern. 

But,  in  1  minute,  the  first  will  empty  -fa  of  the  cistern ;  the 
second  -pfa,  and  the  third  -^ ;  and  together  they  will  empty 
iro  +  irio-  +  iroir  of  tne  cistern. 

Therefore  —  +  —  +  —  =  -• 

80     200     300     x 

Whence  x  —  48. 

It  will  take  the  three  taps  together  48  minutes,  or  £•  of  an 
hour,  to  empty  the  cistern. 


164  ALGEBRA.  [Cn.  VIII 

EXERCISES  II. 

1.  What  number  added  to  the  numerator  and  denominator 
of  f  will  give  a  fraction  equal  to  f  ? 

2.  The  sum  of  two  numbers  is  18,  and  the  quotient  of  the 
less  divided   by   the   greater  is   equal   to  ^.     What  are  the 
numbers  ? 

3.  The  denominator  of  a  fraction  exceeds  its  numerator  by 
2,  and  if  1  be  added  to  both  numerator  and  denominator,  the 
resulting  fraction  will  be  equal  to  f .     What  is  the  fraction  ? 

4.  The  sum  of  a  number  and  7  times  its  reciprocal  is  8. 
What  is  the  number  ? 

5.  The  value  of  a  fraction,  when  reduced  to  its  lowest 
terms,  is   f.      If   its   numerator  be   increased   by  7   and   its 
denominator  be  decreased  by  7,  the  resulting  fraction  will  be 
equal  to  f .     What  is  the  fraction  ? 

6.  What  number   must  be   added   to  the  numerator   and 
subtracted  from  the  denominator  of  the  fraction  T75,  to  give  its 
reciprocal ? 

7.  If  \  be  divided  by  a  certain  number  increased  by  },  and 
i  be  subtracted  from  the  quotient,  the  remainder  will  be  J. 
What  is  the  number  ? 

8.  A  train  runs  200  miles  in  a  certain  time.     If  it  were  to 
run  5  miles  an  hour  faster,  it  would  run  40  miles  farther  in 
the  same  time.     What  is  the  rate  of  the  train  ? 

9.  A  number  has  three  digits,  which  increase  by  1  from 
left  to  right.     The  quotient  of  the  number  divided  by  the  sum 
of  the  digits  is  26.     What  is  the  number  ? 

10.  A  number  of  men  have  $  72  to  divide.     If  $  144  were 
divided  among  3  more  men,  each  one  would  receive  $4  more. 
How  many  men  are  there  ? 

11.  It  was  intended  to  divide  J  by  a  certain  number,  but  by 
mistake  \  was  added  to  the  number.     The  result  was,  never- 
theless, the  same.     What  is  the  number  ? 


3]  FRACTIONAL  EQUATIONS.  165 

12.  A  steamer  can  run  20  miles  an  hour  in  still  water.     If 
it  can  run  72  miles  with  the  current  in  the  same  time  that  it 
can  run  48  miles  against  the  current,  what  is  the  speed  of  the 
current  ? 

13.  A  man  buys  two  kinds  of  wine,  14  bottles  in  all,  paying 
$  9  for  one  kind  and  $  12  for  the  other.     If  the  price  of  each 
kind  is  the  same,  how  many  bottles  of  each  does  he  buy  ? 

14.  A  farmer  intended  to  feed   80   bushels   of  corn  to   a 
certain  number  of  sheep.     When  6  of  the  sheep  died,  he  could 
have  sold  24  bushels  of  corn  and  have  had  enough  left  to 
give  each  remaining  sheep  the  same  amount  as  before.     How 
many  sheep  had  he  ? 

15.  It  takes  a  pedestrian  5  hours  to  go  from  A  to  B.     It 
takes  a  bicycle  rider,  who  goes  6  miles  farther  every  hour, 
2  hours  to  go  the  same  distance.     How  far  is  A  from  B  ? 

16.  A  can  do  a  piece  of  work  in  10  days,  B  in  6  days  and 
A,  B,  and  C  together  in  3  days.     In  how  many  days  can  C  do 
the  work  ? 

17.  A  and  B  together  can  do  a  piece  of  work  in  2  days,  B 
and  C  together  in  3  days,  and  A  and  C  together  in  21  days. 
In  how  many  days  can  A,  B,  and  C  together  do  the  work  ? 

18.  The   circumference   of   the  hind   wheel   of   a  carriage 
exceeds  the  circumference  of  the  front  wheel  by  4  feet,  and 
the  front  wheel  makes  the  same  number  of  revolutions  in  run- 
ning 400  yards  that  the  hind  wheel  makes  in  running  500  yards. 
What  is  the  circumference  of  each  wheel  ? 

19.  A  cistern  has  3  taps.     By  the  first  it  can  be  filled  in 
6  hours,  by  the  second  in  8  hours,  and  by  the  third  it  can  -be 
emptied  in  12  hours.     In  what  time  will  it  be  filled  if  all  the 
taps  are  opened  ? 

20.  An  inlet  pipe  can  fill  a  cistern  in  3  hours,  and  an  outlet 
pipe  can  empty  it  in  9  hours.     After  how  many  hours  will  the 
cistern  be  filled  if  both  pipes  are  open  one-half  of  the  time, 
and  the  outlet  pipe  is  closed  during  the  second  half  of  the 
time  ? 


166  ALGEBRA.  [Cn.  VIII 

21.  In  a  number  of  two  digits,  the  digit  in  the  tens'  place 
•exceeds  the  digit  in  the  units'  place  by  2.     If  the  digits  be 
interchanged   and   the   resulting  number   be   divided   by   the 
original  number,  the  quotient  will  be  equal  to  |~|.     What  is 
the  number  ? 

22.  In  a  number  of  three  digits,  the  digit  in  the  hundreds' 
place  is  2 ;  if  this  digit  be  transferred  to  the  units'  place,  and 
the  resulting  number  be  divided  by  the  original  number,  the 
quotient  will  be  equal  to  J^.     What  is  the  number  ? 

23.  In  one  hour  a  train  runs  10  miles  farther  than  a  man 
rides  on  a  bicycle  in  the  same  time.     If  it  takes  the  train 
€  hours  longer  to  run  255  miles  than  it  takes  the  man  to  ride 
€3  miles,  what  is  the  rate  of  the  train  ? 

24.  A  cistern  has  three  pipes.     To  fill  it,  the  first  pipe  takes 
one-half  of  the  time  required  by  the  second,  and  the  second 
takes  two-thirds  of  the  time  required  by  the  third.     If  the 
three   pipes   be  open   together,  the   cistern  will   be  filled  in 
<5  hours.     In  what  time  will  each  pipe  fill  the  cistern  ? 

25.  A   and  B  ride   100   miles   from   P  to    Q.     They   ride 
together  at  a  uniform  rate  until  they  are  within  30  miles  of 
•Q,  when  A  increases  his  rate  by  ^  of  his  previous  rate.     When 
B  is  within  20  miles  of  Q,  he  increases  his  rate  by  ^  of  his 
previous  rate,  and  arrives  at  Q  10  minutes  earlier  than  A.     At 
what  rate  did  A  and  B  first  ride  ? 

26.  A  circular  road  has  three  stations,  A,  jB,  and   C,   so 
placed  that  A  is  15  miles  from  B,  B  is  13  miles  from  C  in  the 
same  direction,  and  C  is  14  miles  from  A  in  the  same  direction. 
Two  messengers  leaving  A  at  the  same  time,  and  travelling 
in  opposite  directions,  meet  at  B.     The  faster  messenger  then 
reaches  A  7  hours  before  the  slower  one.     What  is  the  rate 
of  each  messenger  ? 


CHAPTER   IX. 

LITERAL   EQUATIONS   IN   ONE   UNKNOWN   NUMBER. 

1.  The  unknown  numbers  of  an  equation  are  frequently  to 
be  determined  in  terms  of  general  numbers,  i.e.,  in  terms  of 
numbers  represented  by  letters.      The   latter  are  commonly 
represented  by  the  leading  letters  of  the  alphabet,  a,  b,  c,  etc. 

Such  numbers  as  a,  b,  c,  etc.,  are  to  be  regarded  as  known. 
E.g.,  in  the  equation  x  -f-  a  =  b,  a  and  b  are  the  known  num- 
bers, and  x  is  the  unknown  number. 

From  this  equation  we  obtain  x  =  b  —  a. 

2.  A  Numerical  Equation  is  one  in  which  all  the  known  num- 
bers are  numerals  ;  as  2  a?  -f-  3  =  7  ;  4  a?  —  3y  =  7. 

A  Literal  Equation  is  one  in  which  some  or  all  of  the  known 
numbers  are  literal  ;  as  2  ax  -f-  3  b  =  5  ;  ax  +  by  =  c. 


3.  Ex.  1.    Solve  the  equation 


b  a  2ab 

Clearing  of  fractions, 

2  ax  -  2  a2  +  2  bx  -  2  b2  =  -  a2  +  2  ab  -  b2. 
Transferring  and  uniting  terms, 


Dividing  by  2  (a  +  b),     x  =  ^ 


Notice  that  the  above  equation,  although  algebraically  frac- 
tional, is  integral  in  the  unknown  number  x.  The  equation 
which  follows  is  fractional  in  the  unknown  number. 

167 


168  ALGEBRA.  [Cn.  IX 

Ex.  2.    Solve  the  equation  ^t_^  =  ^L±JL 

b  +  x     6  +  1 

Multiplying  by  (&+»)(&  +  !),  (a+»)(&  +  l)  =  (&+aj)  («  +  !)• 
Simplifying,  a&  +  60;  +  a  +  x  =  ab  +  ax + 6  +  a;. 

Cancelling  terms,  &e  +  a  =  ax  -f-  6. 

Transferring  and  uniting  terms,       (b  —  a)x  =  b  —  a. 
Dividing  by  b  —  a,  x  =  1. 

EXERCISES  I. 

Solve  the  following  equations  : 
1.   a  —  x  =  c.  2.   mx  +  a,  =  b.  3.   rax  =  wic  +  2. 

4.  3  ax  —  5  a&  +  6  ax  —  7  ac  =  2  ax  +  2  ab. 

5.  4  a2  -  2  afrc  +  62  +  3  a?x  =  5a?-b2x  +  2  a?x. 

6.  a  (a  H-  a)  -  b(x  -  b)  =  Sax  +  (a  -  6)2. 

7.  05  (x  +  a)  +  a;  (x  +  6)  —  2  (a?  +  a)  (x  +  6)  =  0. 

,6 

8.    a  -f-  -  =  c. 


x  b     x  —  a2 

10   ^JL^  =  5.  11   ^l_i_5_^  =  ^. 

a  —  a     4  ax     a     b     x 

x  +  a         2     _  x  —  a 


12              =           t  13 

b  +  x     6  +  1*  2         «  +  a 

6x-f«      ^a;~~^_o  15    «-  +  ^  _  «  —  x 

4:X+b     2x  —  a  '   b  +  a     b—a 


^ 
'  ' 


x-ab     a2-ab  +  b2  x  -  b      (2x-b)2 

18      ^2  +  a"  x  _1. 

'   4JB2  —  a2     2x  +  a         4* 


' 


3-4]  LITERAL   EQUATIONS.  169 

x-a     x-b      a?-c  =  l  .1.1. 
2bc        2ac        2ab      a     be 

l-2ax2     1+200*         4  aba? 

22.     - — 


23. 


2bx2 
g 


a?  -\-  x  —  a2  +  a     x  +  a     a;  —  a  +  1 
24     a2  +  a;  _  a2 -a;  =  4  abx  +  2  a2  -  2  6^ 


25          <*2  +  a» 


+  a~a?  +  aar  +  or     a4  +  2  a^ar  +  a;4     a  +  a? 
2 /£      2  a  +  a?  a4 


a2  —  a?      o?x  +  2  aa;  —  2  a3  — 


—  a 


General  Problems. 

4.  A  General  Problem  is  one  in  which  the  known  numbers  are 
literal. 

Pr.  1.  The  greater  of  two  numbers  is  m  times  the  less,  and 
their  sum  is  s.  What  are  the  numbers  ? 

Let  x  stand  for  the  less  required  number.  Then  mx  stands 
for  the  greater.  By  the  condition  of  the  problem,  we  have 

x  -f  mx  =  s  ; 

whence,  x—  —  -  —  ,  the  less  number,  and  mx=    ms  ,  the  greater. 
1+m  1+m 

If  m  =  3  and  s  =  84,  we  have 

x  =  J*i-  =  21,  and  mx  =  3  x  21  =  63. 
1  +  3 

When  the  numbers  are  equal,  m  =  1,  and  we  obtain 
aj  =   ,  and  ma?=, 


170  ALGEBRA.  [Cn.  TX 

for  all  values  of  s ;  that  is,  either  of  the  two  numbers  is  half 
their  sum. 

Thus  the  solution  of  this  general  problem  includes  the  solu- 
tions of  all  like  problems.  A  solution  for  any  like  problem 
is  obtained  by  substituting  particular  values  for  m  and  s,  as 
above. 

Pr.  2.  A  cistern  has  two  taps.  By  the  first  it  can  be  filled 
in  a  minutes,  and  by  the  second  in  b  minutes.  How  many 
minutes  will  it  take  the  two  taps  together  to  fill  the  cistern  ? 

Let  x  stand  for  the  number  of  minutes  it  takes  the  two  taps 
to  fill  the  cistern.  Then,  in  1  minute,  the  two  together  will  fill 

-  of  the  cistern. 
x 

But,  in  1  minute,  the  first  will  fill  -  of  the  cistern,  the  sec- 
ond -;  and  together  they  will  fill  -  +  -  of  the  cistern. 
b  a     b 

Therefore  1  +  1  =  1. 

a     b     x 

Whence  ab 


a  +  b 

This  solution  gives  a  general  rule  for  solving  problems  of 
like  character.  In  a  particular  example,  a  may  be  the  number 
of  minutes  it  takes  a  tap  to  fill  a  cistern,  the  number  of  hours 
it  takes  a  man  to  build  a  wall,  to  dig  a  ditch,  to  plough  a  fieldj 
etc. 

Pr.  3.  If  one  man  can  dig  a  ditch  in  6  days,  and  a  second 
man  in  3  days,  in  how  many  days  can  they  dig  the  ditch, 
working  together  ? 

Substituting  a  =  6,  b  —  3,  in  the  result  of  Pr.  2,  we  have 

6x3  0 

Therefore  they  can  together  dig  the  ditch  in  2  days. 


4]  LITERAL   EQUATIONS.  171 

EXERCISES   II. 

Find  the  general  solution  of  each  of  the  following  problems, 
and  from  this  solution  obtain  the  particular  solution  for  the 
numerical  values  assigned  to  the  literal  numbers  in  the  problem. 

1.  Find  a  number,  such  that  the  result  of  adding  it  to  n  shall 
be  equal  to  n  times  the  number.     Let  n  =  2 ;  5. 

2.  Divide  a  into  two  parts,  such  that  —  of  the  first,  plus  -  of 

m  u- 

the  second,  shall  be  equal  to  b.     Let  a  =  100,  b  =  30,  m  =  3, 
n  =  5. 

3.  A  sum  of  d  dollars  is  divided  between  A  and   B.     B 
receives  b  dollars  as  often  as  A  receives  a  dollars.     How  much 
does  each  receive  ?     Let  d  —  7000,  a  =  3,  b  =  2. 

4.  A  father's  age  exceeds  his  son's  age  by  m  years,  and  the 
sum  of  their  ages  is  n  times  the  son's  age.     What  are  their 
ages  ?     Let  m  =  20,  n  =  4 ;  m  =  25,  n  =  7. 

5.  A  farmer  can  plough  a  field  in  a  days,  and  his  son  in  b 
days ;  in  how  many  days  can  they  plough  the  field,  working 
together  ?     Let  a  =  10,  b  =  15. 

6.  What  time  is  it,  if  the  number  of  hours  which  have 
elapsed  since  noon  is  m  times  the  number  of  hours  to  mid- 
night ?     Let  m  =  \. 

7.  One  pipe  can  fill  a  cistern  in  a  hours,  a  second  in  b  hours, 
and  a  third  in  c  hours.     In  how  many  hours  can  the  three  pipes 
fill  the  cistern,  working  together  ?     Let  a  =  2,  b  =  3,  c  =  6. 

8.  One  pipe  can  fill  a  cistern  in  m  hours,  a  second  in  n  hours, 
and  a  third  can  empty  it  in  p  hours.      After  how  many  hours 
will  the  cistern  be  filled,  if  all  pipes  are  open  ?     Let  m  —  4, 
n  =  6,  p  =  3. 

9.  Two  couriers  start  at  the  same  time  and  move  in  the 
same  direction,  the  first  from  a  place  d  miles  ahead  of   the 
second.     The  first  courier  travels  at  the  rate  of  m^  miles  an 
hour,  and  the  second  at  the  rate  of  wa  miles  an  hour.     After 


172  ALGEBRA.  [Cn.  IX 

how  many  hours  will  the  second  courier  overtake  the  first? 
Let  d  =  15,  mi  =  17,  m2  =  20. 

From  the  result  of  the  preceding  example  find  the  results  of 
Exx.  10-12. 

10.  At  what  rate  must  the  second  courier  travel  in  order  to 
overtake  the  first  after  h  hours  ?     Let  d  =  18,  ?%  =  15,  h  =  3. 

11.  At  what  rate  must  the  first  courier  travel  in  order  that 
the  second   courier   may  overtake  him  after   h  hours  ?     Let 
d  =  12,  m2  =  22,h  =  3. 

12.  How  many  miles  behind  the  first  courier  must  the  second 
start  in  order  to  overtake  the  first  after  h  hours  ?     Let  m1  =  18, 
m2  =  21,  h  =  4. 

13.  In  a  company  are  a  men  and  6  women;  and  to  every 
m  unmarried  men  there  are  n  unmarried  women.     H<3w  many 
married  couples  are   in  the  company  ?     Let  a  =  13,  b  =  17, 
m  =  3,  n  =  5. 

INTERPRETATION   OF  THE   SOLUTIONS  OF  PROBLEMS. 

5.  In  solving  equations  we  do  not  concern  ourselves  with 
the  meaning  of  the  results.     When,  however,  an  equation  has 
arisen  in  connection  with  a  problem,  the  interpretation  of  the 
result  becomes  important.     In  this  chapter  we  shall  interpret 
the  solutions  of  some  linear  equations  in  connection  with  the 
problems  from  which  they  arise. 

Positive  Solutions. 

6.  Pr.    A  company  of  20  people,  men  and  women,  proposed 
to  arrange  a  fair  for  the  benefit  of  a  poor  family.     Each  man 
contributed  $  3,  and  each  woman  $  1.    If  $  55  were  contributed, 
how  many  men  and  how  many  women  were  in  the  company  ? 

Let  x  stand  for  the  number  of  men;  then  the  number  of 
women  was  20  —  x.  The  amount  contributed  by  the  men  was 
3  x  dollars,  that  by  the  women  20  —  x  dollars.  By  the  condition 
of  the  problem,  we  have 

3  x  +  (20  —  x)  =  55 ;  whence  x  =  17|. 


4-8]  INTERPRETATION"   OF   SOLUTIONS.  173 

The  result,  17^,  satisfies  the  equation,  but  not  the  problem. 
For  the  number  of  men  must  be  an  integer.  This  implied  con- 
dition could  not  be  introduced  into  the  equation. 

The  conditions  stated  in  the  problem  are  impossible,  since 
they  are  inconsistent  with  the  implied  condition. 

Negative   Solutions. 

7.  Pr.    A  father  is  40  years  old,  and  his  son  10  years  old. 
After  how  many  years  will  the  father  be  seven  times  as  old  as 
his  son  ? 

Let  x  stand  for  the  required  number  of  years.  Then  after 
x  years  the  father  will  be  40  +  x  years  old,  and  the  son  10  +  x 
years  old.  By  the  condition  of  the  problem,  we  have 

40  +  x  =  7  (10  +  a?),  whence  x  =  -  5.  (1) 

This  result  satisfies  the  equation,  but  not  the  condition  of 
the  problem.  For  since  the  question  of  the  problem  is  "  after 
how  many  years  ?  "  the  result,  if  added  to  the  number  of  years 
in  the  ages  of  father  and  son,  should  increase  them,  and  there- 
fore be  positive.  Consequently,  at  no  time  in  the  future  will 
the  father  be  seven  times  as  old  as  his  son.  But  since  to 
add  —  5  is  equivalent  to  subtracting  5,  we  conclude  that  the 
question  of  the  problem  should  have  been,  "  How  many  years 
ago?" 

The  equation  of  the  problem,  with  this  modified  question,  is : 

40  —  x  =  7  (10  —  x)  ;  whence  x  =  5.  (2) 

Notice  that  equation  (2)  could  have  been  obtained  from  equa- 
tion (1)  by  changing  x  into  —  x. 

8.  The  interpretation  of  a  negative  result  in  a  given  problem 
is  often  facilitated  by  the  following  principle : 

If  —  x  be  substituted  for  x  in  an  equation  which  has  a  negative 
root,  the  resulting  equation  will  have  a  positive  root  of  the  same 
absolute  value  ;  and  vice  versa. 

E.g.,  the   equation  x  +  1=  —  x  —  3  has  the  root  —  2 ; 
while  the  equation  —  x  +  \  =  x  —  3  has  the  root  2. 


174  ALGEBRA.  [Cn.  IX 

9.  Pr.  Two  pocket-books   contain   together  $100.     If  one- 
half  of  the  contents  of  one  pocket-book  and  one-third  of  the 
contents  of  the  other  be  removed,  the  amount  of  money  left  in 
both  will  be  $70.     How  many  dollars  does  each  pocket-book 
contain  ? 

Let  x  stand  for  the  number  of  dollars  contained  in  the  first 
pocket-book ;  then  the  number  of  dollars  contained  in  the 
second  is  100  —  x.  When  one-half  of  the  contents  of  the  first 
and  one-third  of  the  contents  of  the  second  are  removed,  the 

number  of  dollars  remaining  in  the  first  is  -x,  and  in  the  second 

2 

|(100  —  x).     By  the  conditions  of  the  problem,  we  have 

$x  +  |(100  -  a;)  =  70,  whence  x  =  -  20. 
Substituting  —  x  for  x  in  the  given  equation,  we  obtain 
-  \x  +  |(100  +  x)  =  70,  or  §(100  +  x)  -  ±x  =  70. 

This  equation  corresponds  to  the  following  conditions : 
If  x  stand  for  the  number  of  dollars  in  one  pocket-book,  then 
100  4-  x  stands  for  the  number  of  dollars  in  the  other ;  that  is, 
one  pocket-book  contains  $  100  more  than  the  other.  The 
second  condition  of  the  problem,  obtained  from  the  equation, 
is :  two-thirds  of  the  contents  of  one  pocket-book  exceeds 
one-half  of  the  contents  of  the  other  by  $  70.  Therefore  the 
modified  problem  reads  as  follows  : 

Two  pocket-books  contain  a  certain  amount  of  money,  and 
one  contains  $100  more  than  the  other.  If  one-third  of  the 
contents  be  removed  from  the  first  pocket-book,  and  one-half 
of  the  contents  from  the  second,  the  first  will  then  contain 
$  70  more  than  the  second.  How  much  money  is  contained  in 
each  pocket-book  ? 

10.  These  problems  show  that  the  required  modification  of 
an  assumption,  question,  or  condition  of  a  problem  which  has 
led  to  a  negative  result,  consists  in  making  the  assumption, 
question,  or  condition  the  opposite  of  what  it  originally  was. 

Thus,  if  a  positive  result  signify  a  distance  toward  the  right 
from  a  certain  point,  a  negative  result  will  signify  a  distance 
toward  the  left  from  the  same  point ;  and  vice  versa;  etc. 


9-11]  INTERPRETATION  OF   SOLUTIONS.  175 

Zero   Solutions. 

11.  A  zero  result  gives  in  some  cases  the  answer  to  the  ques- 
tion j  in  other  cases  it  proves  its  impossibility. 

Pr.  A  merchant  has  two  kinds  of  wine,  one  worth  $7.25  a 
gallon,  and  the  other  $5.50  a  gallon.  How  many  gallons  of 
each  kind  must  be  taken  to  make  a  mixture  of  16  gallons 
worth  $88? 

Let  x  stand  for  the  number  of  gallons  of  the  first  kind ;  then 
16  —  x  will  stand  for  the  number  of  gallons  of  the  second  kind. 

Therefore,  by  the  condition  of  the  problem,  we  have 

7.25x  +  5.5(16  -  x)  =  88 ;  whence  x  =  0. 

That  is,  no  mixture  which  contains  the  first  kind  of  wine  can 
be  made  to  satisfy  the  condition.  In  fact,  16  gallons  of  the 
second  kind  are  worth  $  88. 

EXERCISE  III. 

Solve  the  following  problems,  and  interpret  the  results. 
Modify  those  problems  which  have  negative  solutions  so  that 
they  will  be  satisfied  by  positive  solutions. 

1.  A  and  B  together  have  $  100.     If  A  spend  one-third  of 
his  share,  and  B  spend  one-fourth  of  his  share,  they  will  then 
have  $  80  left.     What  are  their  respective  shares  ? 

2.  A  father  is  40  years  old,  and  his  son  is  13  years  old ;  after 
how  many  years  will  the  father  be  four  times  as  old  as  his  son  ? 

3.  The  sum  of  the  first  and  third  of  three  consecutive  numbers 
is  equal  to  three  times  the  second.     What  are  the  numbers  ? 

4.  In  a  number  of  two  digits,  the  tens'  digit  is  two-thirds  of 
units'  digit.    If  the  digits  be  interchanged,  the  resulting  number 
will  exceed  the  original  number  by  36.     What  is  the  number  ? 

5.  A  teacher  proposes  30  problems  to  a  pupil.     The  latter  is 
to  receive  8  marks  in  his  favor  for  each  problem  solved,  and  12 
marks  against  him  for  each  problem  not  solved.     If  the  number 
of  marks  against  him  exceed  those  in  his  favor  by  420,  how 
many  problems  will  he  have  solved  ? 


176  ALGEBRA.  [Cn.  IX 

6.  In  a  number  of  two  digits  the  tens'  digit  is  twice  the 
units'    digit.      If   the    digits   be   interchanged,    the    resulting 
number  will  exceed  the  original  number  by  18.     What  is  the 
number  ? 

7.  A  has  $  100,  and  B  has  $  30.     A  spends  twice  as  much 
money  as  B,  and  then  has  left  three  times  as  much  as  B.     How 
much  does  each  one  spend  ? 

Discuss  the  solutions  of  the  following  general  problems. 
State  under  what  conditions  each  solution  is  positive,  negative, 
or  zero.  Also,  in  each  problem,  assign  a  set  of  particular  values 
to  the  general  numbers  which  will  give  an  admissible  solution. 

8.  A  father  is  a  years  old,  and  his  son  is  b  years  old.     After 
how  many  years  will  the  father  be  n  times  as  old  as  his  son  ? 

9.  Having  two  kinds  of  wine  worth  a  and  b  dollars  a  gallon, 
respectively,  how  many  gallons  of  each  kind  must  be  taken  to 
make  a  mixture  of  n  gallons  worth  c  dollars  a  gallon  ? 

10.  Two  couriers,  A  and  B,  start  at  the  same  time  from  two 
stations,  distant  d  miles  from  each  other,  and  travel  in  the  same 
direction.     A  travels   n  times  as  fast  as  B.     Where  will  A 
overtake  B  ? 


CHAPTER   X. 

SIMULTANEOUS   LINEAR   EQUATIONS. 

SYSTEMS   OF   EQUATIONS. 

1.   If  the  linear  equation  in  two  unknown  numbers 

5  (1) 


be  solved  for  y,  we  obtain 

y  =  5  —  x. 

We  may  substitute  in  this  equation  any  particular  numerical 
value  for  oj,  and  obtain  a  corresponding  value  for  y.  Thus, 

when  x  =  1,  y  =  4  ;  when  x  =  2,  /  =  3  ;  when  aj  =  3,  y  —  2  ;  etc. 

In  like  manner  the  equation  could  have  been  solved  for  x  in 
terms  of  y,  and  corresponding  sets  of  values  obtained. 

Any  set  of  corresponding  values  of  x  and  y  satisfies  the  given 
equation,  and  is  therefore  a  solution. 

2.   Solving  the  equation 

y-*  =  l  <2> 

for  y,  we  have  ?/  =  1  -f-  a?.     Then, 

when  x  =  1,  2/  =  2  ;  when  jr  =  2,  /  =  3  ;  when  #  =  3,  2/  =  4  ;  etc. 

Now,  observe  that  equations  (1)  and  (2)  have  the  common 
solution,  a;  =  2,  y  =  3.  It  seems  evident,  and  it  is  proved  in 
School  Algebra,  that  these  equations  have  only  this  solution 
in  common. 

Equations  (1)  and  (2)  express  different  relations  between  the 
unknown  numbers,  and  are  called  Independent  Equations. 

Also,  since  they  are  satisfied  by  a  common  set  of  values  of 
the  unknown  numbers,  they  are  called  Consistent  Equations. 

177 


178  ALGEBRA.  [Cn.  X 

3.  A  System  of  Simultaneous  Equations  is  a  group  of  equa- 
tions which  are  to  be  satisfied  by  the  same  set,  or  sets,  of 
values  of  the  unknown  numbers. 

A  Solution  of  a  system  of  simultaneous  equations  is  a  set 
of  values  of  the  unknown  numbers  which  satisfies  all  of  the 
equations. 

4.  The  examples  of  Arts.  1-2  are  illustrations  of  the  follow- 
ing general  principles  : 

A  system  of  linear  equations  has  a  definite  number  of  solutions. 
(i.)    When  the  number  of  equations  is  the  same  as  the  number 
of  unknown  numbers. 

(ii.)    When  the  equations  are  independent  and  consistent. 

5.  Two  systems  of  equations  are  equivalent  when  every  solution 
of  either  system  is  a  solution  of  the  other. 

E.g.)  the  systems  (I.)  and  (II.)  : 


2x-2y  = 

are  equivalent.     For  they  are  both  satisfied  by  the  solution, 
x  =  2,  y  =  1,  and,  as  we  shall  see  later,  by  no  other  solution. 

6.  If  the  equations          x  -f-  y  =  7, 

*  -  y  =  1, 
be  added,  we  obtain  2  x  =  8, 

in  which  the  uriknown  number  y  does  not  appear.     We  say 
that  y  has  been  eliminated  from  the  given  equations. 

7.  Elimination  is  the  process  of  deriving  from  tw"o  or  more 
equations  an  equation  which  has  one  less  unknown  number. 

Elimination  by  Addition  and  Subtraction. 

8.  Ex.  1.    Solve  the  system  3  x  +  4  y  =  24,  (1) 

5x-6y  =  2.  (2) 

To  eliminate  y,  we  multiply  the  equations  by  such  numbers 
as  will  make  the  coefficients  of  //  numerically  equal. 


3-9]  SIMULTANEOUS  LINEAR    EQUATIONS.  179 

Multiplying  (1)  by  3,        9  x  +  12  y  =  72.  (3) 

Multiplying  (2)  by  2,      10  x  -  12  y  =  4.  (4) 

Adding  (3)  and  (4),  19  x  =  76. 

Whence  x  —  4. 

Substituting  4 for  x  in  (1),  12  +  4y  =  24. 

Whence  y  =  3. 

It  is  proved  in  School  Algebra,  Ch.  XIII.,  that  the  above 
method  is  based  upon  equivalent  equations. 

Consequently  the  required  solution  is  x  =  4,  y  =  3. 

This  solution  may  be  written  4,  3,  it  being  understood  that 
the  first  number  is  the  value  of  x,  and  the  second  the  value 
of  y. 

Ex.  2.    Solve  the  system  12x  +  15y  =  8.  (1) 

16o;  +  9y  =  7.  (2) 

We  will  first  eliminate  x. 

Multiplying  (1)  by  4,      48  a?  +  60  y  =  32.  (3) 

Multiplying  (2)  by  3,      48  x  +  27  y  =  21.  (4) 

Subtracting  (4)  from  (3),  33  y  =  11. 

Whence  y  =  -J-. 

Substituting  i  for  y  in  (1),  12  a;  +  5  =  8. 

Whence  x  =  i. 

Consequently  the  required  solution  is  \,  ^. 

9.  The  examples  of  the  preceding  article  illustrate  the  fol- 
lowing method  of  elimination  by  addition  and  subtraction : 

Multiply  both  members  of  the  equations  by  such  numbers  as 
will  make  the  coefficients  of  one  of  the  unknown  numbers  numeri- 
cally equal.  Subtract,  or  add,  corresponding  members  of  the 
resulting  equations,  and  equate  the  results. 

/Solve  this  equation  in  one  unknown  number.  Substitute  the 
value  of  this  unknown  number  in  the  simpler  of  the  given  equa- 
tions, and  solve  for  the  other  unknown  number. 

The  multipliers  are  obtained  by  dividing  the  L.  C.  M.  of  the 
coefficients  of  the  unknown  number  to  be  eliminated  by  the 
coefficients  of  this  unknown  number.  It  is  better  to  eliminate 
that  unknown  number  which  requires  the  smallest  multipliers. 


180  ALGEBRA.  [Cn.  X 


EXERCISES  I. 

Solve  the  following  systems  of  equations  by  the  method  of 
addition  and  subtraction  : 

!  ,     .  -       ,       ~  ,  » +  y  =  a> 


x-y  =  7.  (x-y  =  b.  U  +  4?/  =  19. 

3s  +  y  =  31,  f4*-7y  =  19, 

15a-22/  =  l5.      '    U  +  9y  =  37. 

nx  —  ay  =  0, 

*'•!-.,  8. 

TTX  —  ay  =  an. 


10.    '  '        "         '  -      -^+^  =  c> 


12. 


16  x  +  9  y  =  7.  I  mx  +  ny  =p. 

f 3  aj  -f  16  y  =  5,  f  21  a?  +  8  y  =  —  66, 

I  -  5  x  4-  28  y  =  19.  13    i  28  x  -  23  y  =  13. 


f  18  a?  -  20  y  =  1, 
"L4*  ll5o;  +  162/  =  9. 


15  a;  -  14  y  =  33,  f  25  x  +  24  w  =  98, 

16.    \  17.    \ 

20  a;  +  21  y  =  —  24.  15  a;  -  16  i/  =  -  2. 


f  40  a;  -  63  y  =  57,  f  15  a?  +  28  y  =  58  a, 

18y  =  -87.  '  Il8aj-36y  =  a. 


18     \  19     -I 

'    I35ic-18v  =  87. 


Elimination  by  Comparison. 
10.  Ex.   Solve  the  system 

7x  +  2y  =  20,  (1) 

13a-32/  =  17.  (2) 

To  eliminate  jc,  w«  proceed  as  follows : 

Solving  (1)  for  y,  y  =  2°~7a;.  (3) 

Solving  (2)  for  y,  =  13^-17,  (4) 

o 

Equating  these  values  of  y, 

20 -7  x     13  a;  -  17  ,~ 
o — o *  (Pi 


9-11]  SIMULTANEOUS  LINEAR  EQUATIONS.  181 

Whence  x  =  2. 

Substituting  2  for  x  in  (3),  y  =  2Q  ~  14  =  3. 

It  is  proved  in  School  Algebra,  Ch.  XIII.,  that  the  above 
method  is  based  upon  equivalent  equations. 
Consequently  the  required  solution  is  2,  3. 

11.  This  example  illustrates  the  following  method  of  elimi- 
nation by  comparison : 

Solve  the  given  equations  for  the  unknown  number  to  be  elimi- 
nated, and  equate  the  expressions  thus  obtained.  The  derived 
equation  will  contain  but  one  unknown  number. 

Solve  this  derived  equation,  and  substitute  the  value  of  the 
unknown  number  thus  obtained  in  the  simplest  of  the  preceding4 
equations,  and  solve  for  the  other  unknown  number. 

EXERCISES   II. 

Solve  the  following  systems  of  equations  by  the  method  of 
comparison : 


=  28,  (21x-23y  =  2,. 


' 


' 


' 


r8«  +  3i/  =  58 
(3a;  -8    =  - 


^  ' 


182  ALGEBRA.  [Cii.  X 

63  a -46  y  =  29, 


.y  +  c(x  +  l)  =  0. 

=  9  a  —  b, 


18. 


Elimination  by  Substitution. 

12.   Ex.     Solve  the  system 

5x-2u  =  l,  (1) 

4a;  +  5y  =  47.  (2) 

If  we  wish  to  eliminate  a;,  we  proceed  as  follows : 

Solving  (1)  for  x,  x  = 1  +  2y.  (3) 

o 

Substituting        "T    ^  for  a?  in  (2), 
o 


•c- 


(4) 


Whence  y  =  7. 

Substituting  7  for  y  in  (3),       a?  =  3. 


It  is  proved  in  School  Algebra,  Ch.  XIII.,  that  the  above 
method  is  based  upon  equivalent  equations. 
Consequently  the  required  solution  is  3,  7. 

13.  This  example  illustrates  the  following  method  of  elimi- 
nation by  substitution  : 

Solve  the  simpler  equation  for  the  unknoivn  number  to  be  elimi- 
nated in  terms  of  the  other.  Substitute  the  value  thus  obtained 
in  the  other  equation.  The  derived  equation  will  contain  but  one 
unknown  number. 

Solve  the  derived  equation,  and  substitute  the  value  of  the 
unknown  number  thus  obtained  in  the  expression  for  the  other 
unknown  number,  and  solve  for  the  other  unknown  number. 


11-14]          SIMULTANEOUS  LINEAR  EQUATIONS.  183 


EXERCISES   III. 

Solve  the  following  systems  of  equations  by  the  method 
of  substitution: 


1.  *• 

=  y.  (2x  =  y. 

x  =  2y-3,  (x  =  3y-7, 


{ 5  x  =  4  y.  { x  —  ny  =  0. 

9.    -I  10 

(8x-5y  =  Q. 


11.       _  12. 


13 


4aj-15y  =  22,  flOa-  21  y  =  75, 

6a?  +  7     =-26. 


Linear  Equations  in  Three  Unknown  Numbers. 

14.  The  following  examples  will  illustrate  a  method  of 
solving  systems  of  three  linear  equations  in  three  unknown 
numbers  : 


Ex.  1.    Solve  the  system  2x  —  3y  +  5z  =  ll,  (1) 

5a  +  42/-6z=-5,  (2) 

—  4#  +  7y  —  8s  =  —  14.  (3) 
To  eliminate  x,  we  proceed  as  follows  : 

Multiplying  (1)  by  5,  Wx-15y  +  25z  =  55.  (4) 

Multiplying  (2)  by  2,  10aj  +  8y-12s  =  -  10.  (5) 


184  ALGEBRA.  [Cn.  X 

Subtracting  (4)  from  (5),  23  y  -  37  z  =  -  65.          (6) 

Multiplying  (1)  by  2,  4  x  -  6  y  -f  10  z  =  22.  (7) 

Adding  (3)  and  (7),  y  +  2  z  =  8.  (8) 

Solving  (6)  and  (8),  y  ==  2. 

2  =  3. 

Substituting  2  for  ?/  and  3  for  3  in  (1),          x  =  1. 
Consequently  the  required  solution  is  1,  2,  3. 

Ex.  2.    Solve  the  system 

ay  —  cz  =  0,  (1) 

z-x  =  -b,  (2) 

ax  +  %  =  a2  +  6  (a  +  c).  (3) 

Notice  that  by  eliminating  3  from  (1)  and  (2)  we  obtain  an 
equation  in  x  and  y,  which  with  equation  (3)  gives  a  system  of 
two  equations  in  the  same  two  unknown  numbers. 

Solving  (2)  for  z,  z  =  x-  b.  (4) 

Substituting  x  —  b  for  z  in  (1), 

ay  —  ex  +  cb  =  0.  (5) 

Multiplying  (3)  by  a,  a?x  +  afo/  =  a3  4-  a26  +  abc.       (6) 

Multiplying  (5)  by  6,       —  6cx  +  c%  =  —  We.  (7) 

Subtracting  (7)  from  (6),     (a2  +  be)  x  =  a3  4-  orb  +  a&c  +  &2c 


&);    (8) 

whence  a?  =  a  +  6. 

Substituting  a-\-b  for  a;  in  (4),          z  =  a. 
Substituting  a  for  z  in  (1),  ?/  =  c. 

15.   These  examples  illustrate  the  following  method  : 

Eliminate  one  of  the  unknown  numbers  from  any  two  of  the 
equations  j  next  eliminate  the  same  unknown  number  from  the 
third  equation  and  either  of  the  other  two.  Two  equations  in 
the  same  two  unknoivn  numbers  are  thus  derived. 


14-15]         SIMULTANEOUS  LINEAR  EQUATIONS. 


185 


Solve  these  equations  for  the  two  unknown  numbers,  and  sub- 
stitute the  values  thus  obtained  in  the  simplest  equation  which 
contains  the  third  unknown  number. 


EXERCISES   IV. 

Solve  the  following  systems  of  equations : 

_}.  3  y  _[_  3  %  —  19?  x  4-  y  4-  z  =  21, 

4  y  =  3  a?.  2.  6  x  =  5  2, 


1. 


3. 


3  =  2*. 

a  =  42 -17, 


5.      x  -f-  z  =  30, 
[     +  z=32. 


32/  +  22  =0. 

15  ?/  4-  4  2  —  7  x  =  —  44, 
2  =  5  y  —  33, 

[  x  =  5  y  —  22. 


x  4-  z  = 


(y 


=  2  a. 


cc  4-  2  =  9. 


8.      3y-z  =  5, 
(3z-x  =  Q. 

(3x  +  2y-±z  =  15, 
10.      5  x  —  3  y  4-  2  2  =  28, 


12. 


14. 


16. 


x  +  25y-    6«  =  - 


9. 


11. 


5  x  =  34. 

—  2  =  c, 

—  y  =  b, 


2x-   ±y+    92  =  28, 
13.  \  7x+    3y-    52  =  3, 


15. 


2a?-7y+  ^2=     3, 
-45, 
-95. 


9^+3^-20  z= 
[  _l3o?+4y-30z= 


35,  f    80;  — 21  y-    92  = 

37,     17.   J  12x-2S y  +  loz  = 
64. 


186  ALGEBRA.  [Cn.  X 

{  ax  +  by  =  ?>2,  f  x  +  y  -f  z  =  a  +  6  +  c, 

18.    ]  fry  +  cz  =  b2  +  c2,  19.    |  aa  =  by, 

[cz-\-  ax  =  c2.  az  =  cy. 


ax  4-  &y  —  cz  =  a2  -f-  &2> 

ax  —  abz  +  62,  21. 


»  +  a?/  +  arz  +  a3  =  0, 

x  +  6y  H-  52z  +  63  =  0, 
x  4-  c?/  +  c2z  +  c8  =  0. 


16.   It  is  frequently  necessary  to  simplify  the  equations  be- 
fore applying  one  of  the  preceding  methods  : 

Ex.  1.    Solve  the  system 


Clearing  (1)  and  (2)  of  fractions, 

28  +  4a-10x  +  5i/  =  602/-  100,  (3) 

4  x  -  3  +  15  y  -  21  ==  108  -  30  x.  (4) 

Transferring  and  uniting  terms, 

6  x  +  55  y  =  128,  (5) 

34  x  +  15  y  =  132.  (6) 

Multiplying  (5)  by  3,      18  x  +  165  y  =  384.  (7) 

Multiplying  (6)  by  11,  374  x  +  165  y  =  1452.  (8) 

Subtracting  (7)  from  (8),  356  x  =  1068  ; 

whence,  x  =  3. 

Substituting  (3)  for  x  in  (5),  18  +  55  y  =  128; 
whence,  y  =  2. 

Consequently,  the  required  solution  is  3,  2. 

17.   Certain  fractional  equations  are  to  be  solved  for  the 
reciprocals  of  one  or  both  of  the  unknown  numbers. 


15-17]          SIMULTANEOUS  LINEAR   EQUATIONS.  187 

Q  £* 

Ex.  2.    Solve  the  system  - — 1 *—-  =  8,  (1) 

2x  —  oy     y  —  2 


2x-3y  '  y-2 

Let  2x  —  3y  =  u,  y  —  2  =  v. 

Then  (1)  and  (2)  become  ?  +  -  =  8,  (3) 

u      v 

7        °. 

1*2*10.  (4) 

U       V 

We  will  solve  this  system  for  -  and  — 

u          v 

Multiplying  (3)  by  3,       -  4-  —  =  24.  (5) 

\b  \J 

Multiplying  (4)  by  5,        ?  +  ~  =  50.  (6) 

26 
Subtracting  (5)  from  (6),         -  =  26. 

Dividing  by  26,  -  =  1,  or  u  =  1.    . 

Substituting  1  for  u  in  (3),  3  +  -  =  8, 

or  -  =  5. 

v 

Dividing  by  5,  -  =  1,  or  v  =  1. 

We  now  have  to  solve  the  system, 

y-2  =  l.  (8) 

From  (8),  y  =  3. 

Substituting  3  for  y  in  (7),  2  x  —  9  =  1, 
or  2  a?  =  10. 

Dividing  by  2,  x  =  5. 

Therefore  the  required  solution  is  5,  3. 


188  ALGEBRA.  [Cn.  X 

Ex.  3.     Solve  the  system,  x  -f-  y  =  xy,     (1)  j 

2x  +  2z  =  xz,     (2)  (I.) 

3y  +  3z  =  yz.     (3)  J 

Observe  that  the  given  equations  are  neither  linear  nor  frac- 
tional. Yet  they  can  be  transformed  so  that  they  will  contain 
only  the  reciprocals  of  »,  y,  and  z. 

Dividing  (1)  by  xy,  (2)  by  xz,  (3)  by  yz,  we  have  : 


We  will  solve  this  system  for  -,  -,  — 

x    y    z 

Multiplying  (4)  by  2,  -  +  -  =  2.  (7) 

y      x 

o      o 

Subtracting  (5)  from  (7),          -  —  -  =  1.  (8) 

Solving  (6)  and  (8)  for  1  and  -,    -  =  —  ,    !  =  --!. 

Substituting  —  for  -  in  (4),          -  =  —  . 
\2i          y  x      \.2i 

Consequently,  a  solution  of  the  given  system  is  ^,  -^  —  12. 

It  is  important  to  notice  that  we  cannot  assume  that  the 
system  (II.)  is  equivalent  to  the  system  (I.),  since  the  equa- 
tions of  (II.)  are  derived  from  the  equations  of  (I.)  by  dividing 
by  expressions  which  contain  the  unknown  numbers. 

But  if  any  solution  of  (I.)  be  lost  by  this  transformation,  it 
is  a  solution  of  the  expressions  (equated  to  0)  by  which  the 
equations  of  (I.)  were  divided  ;  that  is,  of 

xy  =  0,  xz  =  0,  yz  =  0.  (III.) 

The  system  (III.)  has  the  solution  0,  0,  0,  and  this  solution 
evidently  satisfies  the  system  (I.). 

We  therefore  conclude  that  the  given  system  has  the  two 
solutions  -1/,  Y,  -  12,  and  0,  0,  0. 


17] 


SIMULTANEOUS   LINEAR   EQUATIONS. 


189 


EXERCISES   V. 

Solve  the  following  systems  of  equations : 


1. 


3. 


7. 


te+SI- 

nil    2x- 

22, 
2. 
=  20. 

-5     -7 

oj-1     3 

2/-l~47 

x  +  3     10 

lly       5  - 

x-7     y- 

2/  +  3     13 

2x  +  7y     x  +  1 

3               2                                  4 
x-7     y-5     g 

A                         n                  J 

4               6 

2x  +  72/     a?  -4-  7     0 

2            3 
2x+l     3y+2     0 

6               3 
3aj_4     42/-1 

5              7                          6 
"  3x-l     7y  +  2     0 

2              5 
5^_9     y_2 

A       "'       /-•           ^x      y. 
4              6 

a;      !      y            1 

7             2 
*       1       y         1 

w  +  1      «  - 

-2-T  +  - 

n  —  1      w  - 
9. 

10. 
11. 

-1     n-1 

y         i 

r  —      J 

m  —  a     m  —  o 

x    \    y     i 

f  i    »2  -  1 

5a;  +  7#  +  2     3x- 

n  —  a     Ti  —  6 

h42/  +  7 

3                       4 

7a:  +  3#  +  4     6a  +  5^  +  7 

4                         5 

cc_3x  +  5f  +  17  =  52/  +  4x±7; 
17                                  3 

22-6?(     5x-7     x  +  1     S.v  +  5 

3               11             6             18 
3»  +  7.y  +  l      2a;-3?/  +  8     0 

K                                                 0                              W> 

5                          o 

5a;-77/H-10      3x  +  2y  +  6     0 

3                          5 

' 


13. 


190 


ALGEBRA. 


[Cn.  X 


14. 


17. 


19. 


22. 


7«--  =  16, 

3      4      -<                     a  l  b 

15. 

x      y                           \  x     y 

J.O. 

3a-?  =  4. 

----2                     6  +  a-»i 

y 

ay                            x      y 

r     3            4        o 

3           #  —  5  y      Q 

x  —  4      y  —  1        ' 

x  +  2  w           3 

ia.  - 

9            2        „ 

J.                 O  I/  —  iC        Q  i 

4(^+2  y)         5 

c  1     1 

11                               1       2      2 

x  +  y5' 

-  +  ~  =  a,                   -H  h  -  =  16, 

z      y                           x     y     z 

1  +  1  =  6,           20. 

i  +  i  =  6,          21.      i  +  ?  +  ?=15, 

a?      2! 

2        £C                                      y       Z        X 

1  +  1  =  7.            - 

11                                 199 
-  +  -  =  c.                    i  +  f  +  f^U. 

a?     y                        (z      x     y 

1,3.4 

-  H  h  -  =  o, 

r3  +  4_8  =  15 

x     y      z 

•?  -f  -    * 

M-?=«, 

5       1  +2-i 

a     y      z 

x     2y     9      6 

~x~l^  =  21' 

4  #      w      » 

EXERCISES  VI. 
MISCELLANEOUS    EXAMPLES. 


Solve  the  following  systems  of  equations  by  the  methods 
given  in  this  chapter : 


1. 


x  +  y  =  z  +  10, 


aj-y  +  l 


2.      az  =  3  (a;  +  z), 
[  a#  *=  4  (a?  -f  2/). 

7  11 

2«-3y 


4. 


=  06. 


10?/-7      10 


SIMULTANEOUS  LINEAR  EQUATIONS.  191 


5. 


7. 


-• 


(x  -[-  a  —  b 

x-b 

2ft 

1 

y-a-b 
b 

y-a     e. 

a 

a?  -f-  ny      n 
lOw 

—  ny 
3 

x  —  a 

cw  +  by  , 

y  +  b 

e     a  +  1 

x  +  ny      ft 
f  ™  _i_  -> 

—  ft?/ 
A       1  4- 

2 

ax±by  +  . 

a                         £ 
8. 
^  __&  +  !.                    a^_, 

0  =  1  + 

.2 

2a  ' 


2a 


a?x  -  b2y  =  0, 

(a2  +  &>  +  («2  -  b2)y  =  a4  +  b*. 
(a  +  6)  x  H-  (a  —  6)  ?/  =  a2  -f  &2, 
(a  _  6)a;  +  (a 


.T?/ 

i                 r. 

n 

X2;      —  h                                    12 

\             •         —  W/j 

111 

• 

111 

13 

2/  +2; 
'  x  +  2/  +  2  =  6, 

03  +  2!  +  U  =  8, 

.  y  +  z  +  w  =  9. 

.»•  "J     2 

1^                     1          -1  y 

a?  —  ?/  -f-  2?  +  u  =  9, 
i/  4-  1  ,  5  a;     Q  l 

15 

5  x          4  y  H-  1     3 
16. 

2x         y 

a 

x  .by  .cz          \T)  i  c 

3x        y 

x+y     y+z 

17.   < 

i 
c 

*)       c       a 

a  +  b         a 
\  y  —  x     a  —  b 

1 

« 

")            C 

y  +  az  =  a?  +  c?. 

y  +  x     a  -{-b7 

192 


ALGEBRA. 


[Cn.  X 


20. 


21. 


19. 

rt(a- 

U(a- 
r     a? 

10            -              ^ 

2  a;  +  3  ?/  —  29  '  7  a  —  8  ?/  +  24 

80                       4  * 
o 

h  &  _  C)aj  +  |(a  -  6  +  c)  y  =  a2  +  (b  -  c)2, 
-  b  +  c)  x  4-  £  (a  +  6  —  c)  y  =  a2  —  (b  —  c)2. 
.V       .2     ,2                   y-6       10        y+6 

ri2-!     a2-! 
22. 

5              3 

Ia2-fl  '  n2+l 

ar—3x    3y—xy 

10 


Problems. 

18.  Pr.  1.  The  sum  of  the  two  digits  of  a  number  is  12.  If 
the  digits  are  interchanged,  the  resulting  number  will  exceed 
the  original  one  by  three-fourths  of  the  original  number. 
What  is  the  number? 

Let  x  stand  for  the  units'  digit,  and  y  for  the  tens'  digit. 

Then  the  original  number  is.Wy  +  x. 

When  the  digits  are  interchanged,  the  resulting  number  is 


(1) 


The  first  condition  of  the  problem  states, 
in  verbal  language  :  the  sum  of  the  digits  is  12  ; 
in  algebraic  language  :        x  +  y  =  12. 

The  second  condition  states, 
in  verbal  language  :    the  resulting  number  minus  the  original 

number  is  equal  to  J  of  the  original  number; 
in  algebraic  language  :  10  x  +  y  —  (Wy  +  x)  =  f  (Wy  +  x).     (2) 

Solving  (1)  and  (2),       x  =  8,  y  =  4. 

Therefore  the  required  number  is  48. 

Pr.  2.  A  tank  can  be  filled  by  two  pipes.  If  the  first  is  left 
open  6  minutes,  and  the  second  7  minutes,  the  tank  will  be 
filled;  or  if  the  first  is  left  open  3  minutes,  and  the  second 
12  minutes,  the  tank  will  be  filled.  In  what  time  can  each 
pipe  fill  the  tank  ? 


17-18]         SIMULTANEOUS   LINEAR   EQUATIONS.  193 

Let  x  stand  for  the  number  of  minutes  it  takes  the  first  pipe 
to  fill  the  tank,  and  y  for  the  number  of  minutes  it  takes  the 
second  pipe.  Let  the  capacity  of  the  tank  be  represented  by  1. 

Then  in  1  minute  the  first  pipe  fills  -  of  the  tank,  and  in 

/•»  X  rr 

6  minutes  -  of  the  tank ;  the  second  pipe  fills  -  of  the  tank 
x  y 

in  7  minutes.     Therefore,  by  the  conditions  of  the  problem, 

6,7      ,      3     12      1 

-  +  -=-L;   —  H —  =  1. 
x      y  x      y 

Whence  x  =  lOi,  y  =  17. 

Pr.  3.  The  sum  of  the  three  digits  of  a  number  is  9.  The 
digit  in  the  hundreds'  place  is  equal  to  one-eighth  of  the  number 
composed  of  the  two  other  digits,  and  the  digit  in  the  units7 
place  is  equal  to  one-eighth  of  the  number  composed  of  the  two 
other  digits.  What  is  the  number  ? 

Let  x  stand  for  the  units'  digit, 

y  for  the  tens'  digit, 

and  z  for  the  hundreds'  digit. 

Then,  by  the  first  condition, 

x  +  y  +  *  =  9.  (1) 

The  number  composed  of  the  tens'  and  units'  digits  is  Wy+x. 
Therefore,  by  the  second  condition, 

z  =  $(10y  +  'x).  (2) 

The  number  composed  of  the  hundreds'  and  tens'  digits  is 
10  z  +  y. 

Therefore,  #  =  1  (10  z  +  ?/).  (3) 

Solving  equations  (l)-(3),  we  obtain, 

a>  =  4,   y  =  2,   2  =  3. 
Therefore,  the  required  number  is  324. 

Pr.  4.  The  report  of  a  cannon  travels  172.21  yards  with  the 
wind  toward  A  in  the  same  time  that  it  travels  167.97  yards 
against  the  wind  toward  B.  Three  seconds  after  it  is  fired  it 
is  heard  at  A  and  B,  which  are  2041.08  yards  apart.  What  is 


194  ALGEBRA.  [On.  X 

the  velocity  of  the  report  in  still  air,  and  what  is  the  velocity 
of  the  wind  ? 

Let  x  stand  for  the  number  of  yards  the  report  travels  a 
second  in  still  air, 
and  y  for  the  number  of  yards  the  wind  travels  a  second. 

Then,  in  1  second  the  report  travels  x  +  y  yards  with  the 
wind  toward  A,  and  x  —  y  yards  against  the  wind  toward  B. 

172  21 

Therefore,  it  takes -  seconds  to  travel  172.21  yards 

x  +  y 

-«  r*r^  C\^ 

toward   A.    and    — - —     seconds    to     travel    167.97    yards 
x-y 

toward  B. 


Consequently,  by  the  first  condition, 
172.21      167.97 


(1) 


x  -\-  y       x  —  y 

In  3  seconds  the  report  travels  3(x  +  y)  yards  to  A,  and 
3  (x  —  y)  yards  to  B. 

Therefore,  by  the  second  condition, 

3(x  +  y)+3(x-y)=  2041.08.  (2) 

Solving  equations  (1)  and  (2),  we  obtain 
x :  =  340.18,   y  =  4.24.' 

Therefore,  the  velocity  of  the  report  in  still  air  is  340.18 
yards  a  second,  and  the  velocity  of  the  wind  is  4.24  yards  a 
second. 

Pr.  5.  Two  boys,  A  and  B,  run  a  race  from  P  to  Q  and  return. 
A,  the  faster  runner,  on  his  return  meets  B  90  feet  from  Q,  and 
reaches  P  3  minutes  ahead  of  B.  If  he  had  run  again  to  Q,  he 
would  have  met  B  at  a  distance  from  P  equal  to  one-sixth  of 
the  distance  from  P  to  Q.  How  far  is  Q  from  P,  and  at  what 
rates  do  A  and  B  run  ? 

Let  x  stand  for  the  number  of  feet  from  P  to  Q, 
y  for  the  number  of  feet  A  runs  in  1  minute, 
z  for  the  number  of  feet  B  runs  in  1  minute. 


18]  SIMULTANEOUS  LINEAR   EQUATIONS.  195 

When  they  first  meet  90  feet  from  Q,  A  has  evidently  run 
x  +  90  feet  in  -     —  minutes,  and  B  has  run  x  —  90  feet  in 

-  minutes. 
Therefore,  90  =        90.  (1) 


A  runs  2  x  feet,  from  P  to  Q  and  return,  in  —  minutes,  and 

2x  y 

B  the  same  distance  in  —  minutes. 

z 

Therefore,  by  the  second  condition, 

Zi  X       Zi  X        Q  /o\ 

-  =  --  O.  (4  ) 

y      * 

13  x 

If  A  had  again  met  B,  he  would  have  run  2x  +  ±xf  =  —  —  , 

6 

feet  in  —  -  minutes,  and  B  would  have  run  2x  —  ±x,  =  —-  ~ 
§y  6  ' 

feet  in  —  —  minutes. 
oz 

Therefore,  by  the  last  condition, 

!!*   Or^  =  i!.  (3) 

6z        6z'         y        z 

Solving  equations  (l)-(3),  we  obtain 


Therefore  the  distance  from  P  to  Q  is  1080  feet ;  A  runs 
13Q1-Q-  feet  a  minute,  and  B  runs  HOy^  feet  a  minute. 

EXERCISES  VII. 

1.  Find  two  numbers  whose  sum  is  19  and  whose  difference 
is  7. 

2.  If  one  number  be  multiplied  by  3  and  another  by  7, 
the  sum  of  the  products  will  be  58 ;  if  the  first  be  multiplied 
by  7  and  the  second  by  3,  the  sum  will  be  42.     What  are  the 
numbers  ? 


196  ALGEBRA.  [Cn.  X 

3.  In  a  meeting  of  48  persons,  a  motion  was  carried  by  a 
majority  of  18.     How  many  persons  voted  for  the  motion  and 
how  many  against  it  ? 

4.  If  one  of  two  numbers  be  divided  by  6  and  the  other  by 
5,  the  sum  of  the  quotients  will  be  52 ;  if  the  first  be  divided 
by  8  and  the  second  by  12,  the  sum  of  the  quotients  will  be  31. 
What  are  the  numbers  ? 

5.  Find  two  numbers,  such  that  if  1  be  subtracted  from  the 
first  and  added  to  the  second,  the  results  will  be  equal ;  while 
if  5  be  subtracted  from  the  first  and  the  second  be  subtracted 
from  5,  these  results  will  also  be  equal. 

6.  If  45  be  subtracted  from  a  number,  the  remainder  will 
be  a  certain  multiple  of  5  j  but  if  the  number  be  subtracted 
from  135,  the  remainder  will  be  the  same  multiple  of  10.  What 
is  the  number,  and  what  multiple  of  5  is  the  first  remainder  ? 

7.  If  1  be  added  to  the  numerator  of  a  fraction,  the  result- 
ing fraction  will  be  equal  to  J;    but  if  1  be  added  to  the 
denominator,  the  resulting  fraction  will  be  equal  to  J-.     What 
is  the  fraction  ? 

8.  A  said  to  B :  "  Give  me  three-fourths  of  your  marbles 
and  I  shall  have  100  marbles."     B  said  to  A  :  "  Give  me  one- 
half  of  your  marbles  and  I  shall  have  100  marbles."     How 
many  marbles  had  A  and  B  ? 

9.  A  bag  contains  white  and  black  balls.     One-half  of  the 
number  of  white  balls  is  equal  to  one-third  of  the  number  of 
black  balls,  and  twice  the  number  of  white  balls  is  6  less  than 
the  total  number  of  balls.     How  many  balls  of  each  color  are 
there  ? 

10.  The   sum   of   two   numbers   is   47.     If   the  greater  be 
divided  by  the  less,  the  quotient  and  the  remainder  will  each 
be  5.     What  are  the  numbers  ? 

11.  A  father  said  to  his  son :  "  After  3  years  I  shall  be  three 
times  as  old  as  you  will  be,  and  7  years  ago  I  was  seven  times 
as  old  as  you  then  were."     What  were  the  ages  of  father  and 
son? 


18]  SIMULTANEOUS  LINEAR  EQUATIONS.  197 

12.  A  merchant  received  from   one   customer  $26  for  10 
yards  of  silk  and  4  yards  of  cloth;  and  from  another  customer 
$  23  for  7  yards  of  silk  and  6  yards  of  cloth  at  the  same  prices. 
What  was  the  price  of  the  silk  and  of  the  cloth  ? 

13.  A  merchant  has  two  kinds  of  wine.     If  he  mix  9  gallons 
of  the  poorer  with  7  gallons  of  the  better,  the  mixture  will  be 
worth  $  1.37^  a  gallon ;  but  if  he  mix  3  gallons  of  the  poorer 
with  5  gallons  of  the  better,  the  mixture  will  be  worth  $1.45  a 
gallon.     What  is  the  price  of  each  kind  of  wine  ? 

14.  A  man  has  a  gold  watch,  a  silver  watch,  and  a  chain. 
The  gold  watch  and  the  chain  cost  seven  times  as  much  as  the 
silver  watch ;   the  cost  of  the  chain  and  half  the  cost  of  the 
silver  watch  is  equal  to  three-tenths  of  the  cost  of  the  gold 
watch  ?     If  the  chain  cost  $  40,  what  was  the  cost  of  each 
watch  ? 

15.  A  and  B  make  a  purchase  for  $  48.     A  gives  all  of  his 
money,   and  B  three-fourths  of  his.     If  A  had  given  three- 
fourths  of  his  money  and  B  all  of  his,  they  would  have  paid 
$  1.50  less.     How  much  money  had  A  and  B  ? 

16.  A  mechanic  and  an  apprentice  together  receive  $  40. 
The  mechanic  works  7  days  and  the  apprentice  12  days ;  and 
the  mechanic  earns  in  3  days  $  7  more  than  the  apprentice 
earns  in  5  days.     What  wages  does  each  receive  ? 

17.  I  have  7  silver  balls  equal  in  weight  and  12  gold  balls 
equal  in  weight.     If  I  place  3  silver  balls  in  one  pan  of  a 
balance  and  5  gold  balls  in  the  other,  I  must  add  to  the  gold 
balls  7  ounces  to  maintain  equilibrium.     If  I  place  in  one  pan  4 
silver  balls  and  in  the  other  7  gold  balls,  the  balance  is  in 
equilibrium.     What  is  the  weight  of  each  gold  and  of  each 
silver  ball  ? 

18.  A  tank  has  two  pumps.     If  the  first  be  worked  2  hours 
and  the  second  3  hours,  1020  cubic  feet  of  water  will  be  dis- 
charged.    But  if  the  first  be  worked  1  hour  and  the  second  2J 
hours,  690  cubic  feet  of  water  will  be  discharged.     How  many 
cubic  feet  of  water  can  each  pump  discharge  in  one  hour  ? 


198  ALGEBRA.  [Cn.  X 

19.  It   was   intended   to   distribute  $  25   among   a  certain  . 
number  of  the  poor,  each  adult  to  receive  $  2.50  and  each 
child  75  cents.     But  it  was  found  that  there  were  3  more 
adults  and  5  more  children  than  was  at  first  supposed.     Each 
adult  was  therefore  given  $  1.75  and   each   child   50   cents. 
How  many  adults  and  how  many  children  were  there  ? 

20.  A  man  ordered  a  wine  merchant  to  fill  two  casks  of 
different  sizes  with  wine,  one  at  $  1.20  and  the  other  at  $  1.50 
a  quart,  paying  $  88.50  for  both  casks  of  wine.     By  mistake 
the  casks  were  interchanged,  so  that  the  purchaser  received 
more  of  the  cheaper  wine  and  less  of  the  dearer.     The  mer- 
chant therefore  returned  to  him  $  1.50.     How  many  quarts 
did  each  cask  hold? 

21.  A  and  B  jointly  contribute  $  10,000  to  a  business.     A 
leaves  his  money  in  the  business  1  year  and  3  months,  and  B 
his  money  2  years  and  11  months.     If  their  profits  are  equal, 
how  much  does  each  contribute  ? 

22.  One  boy  said  to  another :  "  Give  me  5  of  your  nuts,  and 
I  shall  have  three  times  as  many  as  you  will  have  left."     "  No," 
said  the  other,  "  give  me  2  of  your  nuts,  and  I  shall  have  five 
times  as  many  as  you  will  have  left."     How  many  nuts  had 
each  boy  ? 

23.  A  father  has  two  sons,  one  4  years  older  than  the  other. 
After  2  years  the  father's  age  will  be  twice  the  joint  ages  of  his 
sons ;  and  6  years  ago  his  age  was  six  times  the  joint  ages  of 
his  sons.     How  old  is  the  father  and  each  of  his  sons  ? 

24.  If  a  number  of  two  digits  be  divided  by  the  sum  of  the 
digits,  the  quotient  will  be  7.     If  the  digits  be  interchanged, 
the  resulting  number  will  be  less  than  the  original  number  by 
27.     What  is  the  number  ? 

25.  A  man  walks  26  miles,  first  at  the  rate  of  3  miles  an 
hour,  and  later  at  the  rate  of  4  miles  an  hour.     If  he  had 
walked  4  miles  an  hour  when  he  walked  3,  and  3  miles  an 
hour  when  he  walked  4,  he  would  have  gone  4  miies  farther. 
How  far  would  he  have  gone,  if  he  had  walked  4  miles  an 
hour  the  whole  time  ? 


18]  SIMULTANEOUS   LINEAR  EQUATIONS.  199 

26.  Two  trains  leave  different  cities,  which  are  650  miles 
apart,  and  run  toward  each  other.     If  they  start  at  the  same 
time,  they  will  meet  after  10  hours ;  but  if  the  first  start  4J- 
hours  earlier  than  the  second,  they  will  meet  8  hours  after  the 
second  train  starts.     What  is  the  speed  of  each  train  ? 

27.  If  the  base  of  a  rectangle  be  increased  by  2  feet,  and 
the  altitude  be  diminished  by  3  feet,  the  area  will  be  dimin- 
ished by  48  square  feet.     But  if  the  base  be  increased  by  3 
feet,  and  the  altitude  be  diminished  by  2  feet,  the  area  will  be 
increased  by  6  square  feet.     Find  the  base  and  the  altitude  of 
the  rectangle. 

28.  A  number  of  three  digits  is  in  value  between  400  and 
500,  and  the  sum  of  its  digits  is  9.     If  the  digits  be  reversed, 
the  resulting  number  will  be  |^-  of  the  original  number.     What 
is  the  number  ? 

29.  The  report  of   a  cannon  travels  with  the  wind  344.42 
yards  a  second,  and  against  the  wind  335.94  yards  a  second. 
What  is  the  velocity  of  the  report  in  still  air,  and  what  is  the 
velocity  of  the  wind  ? 

30.  The  sum  of  three  digits  of  a  number  is  14 ;  the  sum  of 
the  first  and  the  third  digit  is  equal  to  the  second;  and  if  the 
digits  in  the  units'  and.  in  the  tens'  place  be  interchanged,  the 
resulting  number  will  be  less  than  the  original  number  by  18. 
What  is  the  number  ? 

31.  The  sum  of  the  ages  of  A,  B,  and  C  is  69  years.     Two 
years  ago  B's  age  was  equal  to  one-half  of  the  sum  of  the  ages 
of  A  and  C,  and  10  years  hence  the  sum  of  the  ages  of  B  and 
C  will  exceed  A's  age  by  31  years.    What  are  the  present  ages 
of  A,  B,  and  C  ? . 

32.  The  total  capacity  of  three  casks  is  1440  quarts.     Two 
of  them  are  full  and  one  is  empty.     To  fill  the  empty  cask  it 
takes  all  the  contents  of  the  first  and  one-fifth  of  the  contents 
of  the  second,  or  the  contents  of  the  second  and  one-third  of  the 
contents  of  the  first.     What  is  the  capacity  of  each  cask  ? 


200  ALGEBRA.  [Cn.  X 

33.  Three  brothers  wished  to  buy  a  house  worth  $  70,000, 
but  none  of  them  had  enough  money.     If  the  oldest  brother 
had  given  the  second  brother  one-third  of  his  money,  or  the 
youngest  brother  one-fourth  of  his  money,  each  of  the  latter 
would  then  have  had  enough  money  to  buy  the  house.    But  the 
oldest  brother  borrowed  one-half  of  the  money  of  the  youngest 
and  bought  the  house.     How  much  money  had  each  brother  ? 

34.  A  father's  age  is  twenty-one  times  the  difference  between 
the  ages  of  his  two  sons.     Six  years  ago  his  age  was  six  times 
the  sum  of  his   sons'  ages,  and  two  years  hence  it  will  be 
twice  the  sum  of  their  ages.     Find  the  ages  of  the  father  and 
his  two  sons. 

35.  Find  the  contents  of  three  vessels  from  the  following 
data :  If  the  first  be  filled  with  water  and  the  second  be  filled 
from  it,  the  first  will  then  contain  two-thirds  of  its  original 
contents ;  if  from  the  first,  when  full,  the  third  be  filled,  the 
first  will   then   contain   five-ninths   of  its   original  contents ; 
finally,  if  from  the  first,  when  full,  the  second  and  third  be 
filled,  the  first  will  then  contain  8  gallons. 

36.  Two   messengers,  A  and  B,  travel  toward  each  other, 
starting  from  two  cities  which  are  805  miles  distant  from  each 
other.     If  A  starts  5|  hours  earlier  than  B,  they  will  meet  6J- 
hours  after  B  starts.     But  if  B  starts  5|  hours  earlier  than  A, 
they  will  meet  5|  hours  after  A  starts.     At  what  rates  do  A 
and  B  travel  ? 

37.  Each  of  two  servants  was  to  receive  $  160,  a  dress,  and  a 
pair  of  shoes  for  one  year's  services.     One  servant  left  after  8 
months,  and  received  the  dress  and  $  106 ;  the  other  servant 
left  after  9£  months,  and  received  a  pair  of  shoes  and  $  142. 
What  was  the  value  of  the  dress,  and  of  the  pair  of  shoes  ? 

38.  On  the  eve  of  a  battle,  one  army  had  5  men  to  every  6 
men  in  the  other.     The  first  army  lost  14,000  men,  and  the 
second  lost  6000  men.     The  first  army  then  had  2  men  to  every 
3  men  in  the  other.     How  many  men  were  there  originally  in 
each  army  ? 


18]  SIMULTANEOUS  LINEAR  EQUATIONS.  201 

39.  If  the  sum  of  two  numbers,  each  of  three  digits,  be  in- 
creased by  1,  the  result  will  be  1000.     If  the  greater  be  placed 
on  the  left  of  the  less,  and  a  decimal  point  be  placed  between 
them,  the  resulting  number  will  be  six  times  the  number  ob- 
tained  by  placing   the   smaller   number   on.   the   left   of   the 
greater,  with  a  decimal  point  between  them.     What  are  the 
numbers  ? 

40.  Three  cities  A,  B,  and  C,  are  situated  at  the  vertices  of 
a  triangle.     The  distance  from  A  to  C  by  way  of  B  is  82  miles, 
from  B  to  A  by  way  of  C  is  97  miles,  and  from  C  to  B  by  way 
of  A  is  89  miles.     How  far  are  A,  B,  and  C  from  one  another  ? 

41.  A  regiment  of  600  soldiers  is  quartered  in  a  four-story 
building.     On  the  first  floor  are  twice  as  many  men  as  are  on 
the  fourth ;  on  the  second  and  third  are  as  many  men  as  are 
on  the  first  and  fourth ;  and  to  every  7  men  on  the  second  there 
are  5  on  the  third.      How  many  men  are  quartered  on  each 
floor  ? 

42.  Four  men  are  to  do  a  piece  of  work.      A  and  B  can  do 
the  work  in  10  days,  A  and  C  in  12  days,  A  and  D  in  20  days, 
and  B,  C,  and  D  in  7-J-  days.     In  how  many  days  can  each  man 
do  the  work,  and  in  how  many  days  can  they  all  together  do 
the  work  ? 

43.  The  year  in  which  printing  was  invented  is  expressed  by 
a  figure  of  four  digits,  whose  sum  is  14.     The  tens'  digit  is 
one-half  of  the  units'  digit,  and  the  hundreds'  digit  is  equal  to 
the  sum  of  the  thousands'  and  the  tens'  digit.     If  the  digits  be 
reversed,  the  resulting  number  will  be  equal  to  the  original 
number  increased  by  4905.   In  what  year  was  printing  invented  ? 

44.  A  vessel  sails  110  miles  with  the  current  and  70  miles 
against  the  current  in  10  hours.     On  a  second  trip,  it  sails 
88  miles  with  the  current  and  84  miles  against  the  current  in 
the  same  time.     How  many  miles  can  the  vessel  sail  in  still 
water  in  one  hour,  and  what  is  the  speed  of  the  current  ? 


202  ALGEBRA.  [Cn.  X 

45.  A  and  B  run  a  race  of  400  yards.     In  the  first  heat  A 
gives  B  a  start  of  20  seconds,  and  wins  by  50  yards.     In  the 
second  heat  A  gives  B  a  start  of  125  yards,  and  wins  by 
5  seconds.     What  is  the  speed  of  each  runner  ? 

46.  A  and  B  formed  a  partnership.      A  invested  $  20,000 
of  his  own  money  and  f  5000  which  he  borrowed ;  B  invested 
$  22,000  of  his  own  money  and  $  8000  which  he  borrowed  at 
the  same  rate  of  interest  as  was  paid  by  A.     At  the  end  of 
a  year,  A's  share  in  the  profits  amounted  to  $  1750  more  than 
the  interest  on  his  $  5000,  and  B's  share  to  $  2000  more  than 
the  interest  on  his  $  8000.     What  rate  per  cent  interest  did 
they  pay,  and  what  rate  per  cent  did  they  realize  on  their 
investments  ? 

47.  Two  bodies  move  along  the  circumference  of  a  circle 
in  the  same  direction  from  two  different  points,  the  shorter 
distance  between  which,  measured   along  the  circumference, 
is  160  feet.     One  body  will  overtake  the  other  in  32  seconds, 
if  they  move  in  one  direction ;  or  in  40  seconds,  if  they  move 
in  the  opposite  direction.     While  the  one  goes  once  around 
the   circumference,   the   distance    passed   over   by   the   other 
exceeds  the  circumference  by  45  feet.     What  is  the  circum- 
ference of  the  circle,  and  at  what  rates  do  the  bodies  move  ? 

48.  A  number  of  workmen,  who  receive  the  same  wages, 
earn  together  a  certain  sum.     Had  there  been  7  more  work- 
men, and  had  each  one  received  25  cents  more,  their  joint 
earnings  would  have  increased  by  $  18.65.     Had  there  been 
4  fewer  workmen,  and  had  each  one  received  15  cents  less, 
their  joint  earnings  would  have  decreased  by  $  9.20.     How 
many  workmen  are  there,  and  how  much  does  each  one  receive  ? 

49.  A  farmer  has  enough  feed  for  his  oxen  to  last  a  certain 
number  or  days.     If  he  were  to  sell  75  oxen,  his  feed  would 
last  20  days  longer.     If,  however,  he  were  to  buy  100  oxen, 
his  feed  would  last  15  days  less.     How  many  oxen  has  he,  and 
for  how  many  days  has  he  enough  feed  ? 


18]  SIMULTANEOUS   LINEAR  EQUATIONS.  203 

50.  An  alloy  of  tin  and  lead,  weighing  40  pounds,  loses  4 
pounds  in  weight  when  immersed  in  water.     Find  the  amount 
of  tin  and  lead  in  the  alloy,  if  10  pounds  of  tin  lose  If  pounds 
when  immersed  in  water,  and  5  pounds  of  lead  lose  .375  of 
a  pound. 

51.  Two  men  were  to  receive  $  96  for  a  certain  piece  of 
work,  which  they  could  do  together  in  30  days.     After  half  of 
the  work  was  done,  one  of  them  stopped  for  8  days,  and  then 
the  other  stopped  for  4  days.     They  finally  completed  the 
work  in  35-J-  days.     How  many  dollars  should  each  one  receive, 
and  in  what  time  could  each  one  have  done  the  work  alone  ? 

52.  It  took  a  certain  number  of  workmen  6  hours  to  carry  a 
pile  of  stones  from  one  place  to  another.     Had  there  been  2 
more  workmen,  and  had  each  one  carried  4  pounds  more  at 
each  trip,  it  would  have  taken  them  1  hour  less  to  complete 
the  work.     Had  there  been  3  fewer  workmen,  and  had  each 
one  carried  5  pounds  less  at  each  trip,  it  would  have  taken 
them  2  hours  longer  to  complete  the  work.     How  many  work- 
men were  there,  and  how  many  pounds  did  each  one  carry  at 
every  trip  ? 

53.  Three  carriages  travel  from  A  to  B.     The  second  carriage 
travels  every  4  hours  1  mile  less  than  the  first,  and  is  4  hours 
longer  in   making  the   journey.     The   third  carriage   travels 
every  3  hours  If  miles  more  than  the  second,  and  is  7  hours 
less  in  making  the  journey.     How  far  is  B  from  A,  and  how 
many  hours  does  it  take  each  carriage  to  make  the  journey  ? 

54:  A  fox  pursued  by  a  dog  is  60  of  her  own  leaps  ahead  of 
the  dog.  The  fox  makes  9  leaps  while  the  dog  makes  6,  but 
the  dog  goes  as  far  in  3  leaps  as  the  fox  goes  in  7.  How  many 
leaps  does  each  make  before  the  dog  catches  the  fox  ? 


CHAPTER   XI. 

INEQUALITIES. 

1.  One  number  is  greater  than  a  second  number  when  the 
remainder  obtained  by  subtracting  the  second  number  from  the 
first  is  positive. 

Thus,  since  6  —  4,  =  2,  is  positive,  6  >  4. 

One  number  is  less  than  a  second  number  when  the  remainder 
obtained  by  subtracting  the  second  number  from  the  first  is 
negative. 

Thus,  since  —  5  —  2,  =  —  7,  is  negative,  —  5  <  —  2. 

In  general, 

a  >  b,  when  a  —  b  is  2^ositive, 

and  a  <  b,  when  a  —  b  is  negative. 

2.  An  Inequality  is  a  statement  that  two  numbers  or  expres- 
sions are  unequal ;  as  ar  +  b2  >  a2. 

The  members  or  sides  of  an  inequality  are  the  numbers  or 
expressions  which  are  connected  by  one  of  the  signs  of  in- 
equality, >  or  <. 

3.  Two  inequalities  are  of  the  Same  or  Opposite  Species,  or 
are  said  to  subsist  in  the  same  or  opposite  sense,  according  as 
they  have  the  same  or  opposite  sign  of  inequality. 

E.g.,  8  >  3  and  —  5  >  —  7  are  inequalities  of  the  same  spe- 
cies ;  0  >  —  1  and  0  <  1  are  inequalities  of  opposite  species. 

Principles  of  Inequalities. 

4.  A  relation  of  inequality  between  two  numbers  can  be 
stated  in  two  ways  ;  as  7  >  3,  or  3  <  7. 

That  is,  if  the  members  of  an  inequality  be  interchanged,  the 
sign  of  inequality  must  be  reversed. 

204 


1-8]  INEQUALITIES.  205 

5.  If  one  number  be  greater  than  a  second,  and  this  second 
number  be  greater  than  a  third,  then  the  first  number  is  greater 
than  the  third;  that  is, 

If  a  >  b  and  b  >  c,  then  a  >  c. 

In  like  manner,  if  a  <  b  and  b  <  c,  then  a  <  c. 

E.g.,  3>2,  2>1,  and  3>1;   -3<-2,   -2<0,  and  -3<0. 

6.  An  inequality  will  continue  to  be  of  the  same  species, 

(i.)    When  the  same  number  is  added  to,  or  subtracted  from, 
each  member. 

(ii.)   When  each  member  is  multiplied  or  divided  by  the  same 
positive  number. 

That  is,  if  a  >  b, 

then  a  +  /?  >  6  +  /i,   a  —  /?  >  6  —  /? ; 

and  an  >  bn,        a  -r-  n  >  b  -r-  n  ; 

wherein  n  is  positive. 

E.g.,  8  >  4,  and  8  +  2  >  4  +  2,   8-2>4-2; 
and  8x2>4x2,   8  -=-  2  >  4  -=-  2. 

7.  An  inequality  will  be  reversed, 

(i.)    When  each  member  is  subtracted  from  the  same  number. 
(ii.)    When  each  member  is  multiplied  or  divided  by  the  same 
negative  number. 

That  is,  if  a  >  b, 

then         n  —  a  <  n  —  b,  a  (—/?)<  6  (—/?), < 


—  n      —  n 

E.g.,          8  >  4,  and  5  -  8  <  5  -  4,  or  -  3  <  1 ; 

8(-2)<4(-2),or-16<-8;  and  _§-<-i-,or  _4<-2. 

8.  There  is  often  an  advantage  in  using  the  same  letter 
with  some  distinguishing  marks  to  represent  different  numbers 
in  the  same  discussion. 

Thus,  with  subscripts:  alt  az,  as,  etc.,  read  a  sub-one,  a  sub- 
two,  a  sub-three,  etc.,  or  simply  a  one,  a  two,  a  three,  etc. 


206  ALGEBRA.  [Cn.  XI 

A  subscript  must  not  be  confused  with  an  exponent.  Thus, 
a3  stands  for  the  product  aaa;  while  a3  is  a  notation  for  a 
single  number. 

Two  or  More  Inequalities. 

9.  If  the  corresponding  members  of  two  or  more  inequalities  of 
the  same  species  be  added,  the  resulting  inequality  will  be  of  the 
same  species. 

That  is,  if  aj  >  61?  a2  >  62,  •••>  then  ax  -f  a2---  >  bl  +  02---. 
E.0.,  -5>-7,  3>2,  and  -o+3>-7  +  2;    or,   -2>-5. 

10.  If  all  the  members  of  two  or  more  inequalities  of  the  same 
species  be  positive,  and  if  the  corresponding  members  be  multiplied 
together,  the  resulting  inequality  will  be  of  the  same  species. 

That  is,  if  ^  >  61?  a2  >  b^  a3  >  63,  then  a^a^  >  MA? 
wherein  a1?  bl}  a2)  b2,  a3,  bB  are  all  positive. 

E.g.,     12  >  4,  3  >  2,  and  12  x  3  >  4  x  2,  or  36  >  8. 

11.  If  the  members  of  one  inequality  be  subtracted  from,  or 
divided  by,  the  corresponding  members  of  another  inequality  of 
the  same  species,  the  resulting  inequality  will  not  necessarily  be  of 
the  same  species. 

That  is,  if  ^  >  ^  and  a2  >  b.,, 

then  Oi  —  a2  may  or  may  not  >  b±  —  b2, 

and  —  may  or  may  not  >  -*• 

a2  02 

E.g.,     11  >  6,  4  >  3,  and  11  -  4  >  6  -  3,  *£  >  f ; 
5  >  4,  3  >  1,   but    5  -  3  <  4  -  1,    |  <  f ; 
8  >  6,  4  >  2,  while  8-4  =  6-2; 

8  >  6,  4  >  3,  while  |  =  f . 
These  examples  show  the  truth  of  the  principle  enunciated. 

12.  Transformation  of   Inequalities.  —  The  preceding  princi- 
ples enable  us  to  make  the  following  transformations  of  in- 
equalities : 


8-13]  INEQUALITIES.  207 

(i.)  Any  term  may  be  transferred  from  one  member  of  an  in- 
equality to  the  other,  if  its  sign  be  reversed. 

E.g.,  if  a  —  b  >  c,  then  a  >  b  +  c. 

(ii.)  If  the  signs  of  both  members  of  an  inequality  be  reversed 
from  -\-to-,  or  from  —  to  +,  the  sign  of  inequality  must  be 
reversed. 

E.g.,  —  3  <  5,  and  3  >  —  5. 

13.  Ex.  1.   Find  one  limit  of  the  values  of  x,  if 

x>  5  x  —  10. 

Transferring  5  a?,  —  4  x  >  —  10. 

Dividing  by  —  4,  x  <  2-J-. 

That  is,  the  inequality  is  satisfied  by  all  values  of  x  less 
than  2£. 

Ex.  2.   Find  the  limits  of  the  values  of  x,  if 

x-5<4-2x,  (1) 

and  5-2a»7-4».  (2) 

Transferring  in  (1),  3  x  <  9,  whence  x  <  3 ; 
Transferring  in  (2),  2  x  >  2,  whence  x  >  1. 
Therefore  the  values  of  x  lie  between  3  and  1. 

Ex.  3.   What  values  of  x  and  y  satisfy  the  inequality 

5x  +  3y>tt,  (1) 

and  the  equality  3  x  -f-  5  £/  =  13  ?  (2) 

Multiplying  (1)  by  3,      15  a?  +  9  y  >  33.  (3) 

Multiplying  (2)  by  5,    15  x  +  25  ?/  =  65.  (4) 
Subtracting  (4)  from  (3),      -  16  y  >  -  32,  or  y  <  2. 

Multiplying  (1)  by  5,    25  x  +  15  y  >  55.  (5) 

Multiplying  (2)  by  3,      9  x  +  15  ?/  =  39.  (6) 
Subtracting  (6)  from  (5),          16  x  >  16,  or  x  >  1. 


208  ALGEBRA.  [Cn.  XI 

Pr.  l.  A  man  receives  from  an  investment  an  integral  num- 
ber of  dollars  a  day.  He  calculates  that  if  he  were  to  receive 
$  6  more  a  day  his  investment  would  yield  over  $  270  a  week  ; 
but  that,  if  he  were  to  receive  $  14  less  a  day,  his  investment 
would  not  yield  as  much  as  $  270  in  two  weeks.  How  much 
does  he  receive  a  day  from  his  investment  ? 

Let  x  stand  for  the  number  of  dollars  which  he  receives  a 
day. 

Then,  by  the  first  condition, 

7  (x  _|_  6)  >  270  ;  whence  x  >  32f 

And,  by  the  second  condition, 

14  (a;  -  14)  <  270  ;  whence  x  <  33f 

Therefore  he  receives  $  33  a  day  from  his  investment. 

EXERCISES   I. 

Determine  one  limit  of  the  value  of  x  in  each  of  the  following 
inequalities  : 

1.   x  _  8  >  4.  2.    -  3  (x  +  10)  >  -  20 

3x  —  8  37  —  2  x     Q       4    11  a  —  x     a  —  x 

~~  ~~~~ 


5.   «5--_eL_<l-l=.  6.   _£_  +  _*_<  2  a. 

1  —  a  a  —  1  a  +  o     a  —  b 

Determine  the  limits  of  the  values  of  x  in  each  of  the  fol- 
lowing systems  of  inequalities  : 

|6x  +  l>0, 
'  ' 


Determine  the  limits  of  the  values  of  x  and  y  in  each  of  the 
following  systems: 

9     (2x  +  3y  =  -±,  1Q    J7a>  +  y  =  15, 

[x-y>  2.  \3x-2y>14:. 

11.  What  integers  have  each  the  property  that  one-half  of 
the  integer,  increased  by  5,  is  greater  than  four-thirds  of  it, 
diminished  by  3  ? 


13-15]  INEQUALITIES.  209 

12.  What  integers  have  each  the  property  that,  if  9  be  sub- 
tracted from  three  times  the  integer,  the  remainder  will  be  less 
than  twice  the  integer,  increased  by  12  ? 

13.  A  has  three  times  as  much  money  as  B.     If  B  gives  A 
$  10,  then  A  will  have  more  than  seven  times  as  much  as  B 
will  have  left.    What  are  the  possible  amounts  of  money  which 
A  and  B  have  ? 

Identical  Inequalities. 

14.  Many  inequalities  hold  for  all  values  of  the  literal  num- 
bers involved ;  as  a2  +  b2  >  a2. 

Such  inequalities  are  analogous  to  identical  equations. 

15.  Prove  that  if  a  is  not  equal  to  6,  then  a2  +  62  >  2  ab. 
We  have  (a  -  6)2  >  0,  (1) 

since  the  square  of  any  positive  or  negative  number  is  positive, 
and  therefore  greater  than  0. 

From  (1),  a2-2ab  +  b2>0; 

whence  a2  +  b2  >  2  ab,  by  Art.  12  (i.). 

EXERCISES   II. 

Prove  the  following  inequalities,  in  which  the  literal  num- 
bers are  all  positive  and  unequal : 

1.  a2  -f  b2  +  c2  >  ab  +  ac  -f  be. 

2.  a2b2  +  b2c2  +  a2c2  >  abc  (a  +  b  +  c). 

3.  ab  (a  -f  b)  +  be  (b  +  c)  -f  ac  (a  +  c)  >  6  abc. 

4.  If  I2  +  m2  +  n2  =  1,  and  I?  +  m?  +  nf  =  1,  then 

Hi  +  wm!  +  nn±  <  1. 

5.  a3  +  bs  >  a2b  +  ab2.  6.   a4  +  64  >  a3&  +  a&3. 

7.  (a  +  6)  (6  -f  c)  (c  +  a)  >  8  a&c. 

8.  3(a2  -f  b2  +  c2)  >  (a  +  b  +  c)2. 


CHAPTER   XII. 

INDETERMINATE   LINEAR   EQUATIONS. 

1.  It  was  shown  in  Ch.  X.,  Art.  1,  that  the  linear  equation  in 
two  unknown  numbers 

is  satisfied  by  an  indefinite  number  of  sets  of  values  of  x  and  y. 

An  Indeterminate  Equation  is  an  equation  which,  like  the 
above,  has  an  indefinite  number  of  solutions. 

Evidently  the  number  of  solutions  will  be  more  limited  if 
only  positive  integral  values  of  the  unknown  numbers  are 
admitted. 

In  this  chapter  we  shall  consider  a  simple  method  of  solving 
in  positive  integers  linear  indeterminate  equations. 

2.  Ex.  1.    Solve  4  x  -\-  7  y  =  94,  in  positive  integers. 
Solving  for  x,  which  has  the  smaller  coefficient,  we  obtain 

(i) 

<±  <± 

or 

O      V      y 

Since  x  and  y  are  to  be  integers, *  must  be  an  integer. 

That  is,  y  must  have  such  a  value  that  2  —  3  y  shall  be  divisi- 
ble by  4. 

2  —  3  v 

Let  ^  =  m,  an  integer. 

4 

Then  y  =  ~~  m,  an  inconvenient  form  from  which  to  de- 
termine integral  values  of  y.  But  since  the  expression  - 

is  to  be  an  integer,  any  multiple  of  it  will  be  an  integer.     We 
therefore  multiply  its  numerator  by  the  least  number  which 

210 


1-4]  INTERMEDIATE   LINEAR  EQUATIONS.  211 

will  make  the  coefficient  of  y  one  more  than  a  multiple  of  the 
denominator,  i.e.,  by  3. 
We  then  have 

*=£y  =l-2y+  2^1,  an  integer. 

2  —  v 

Therefore,  -  -  =  m.  an  integer. 

4 

Whence  y  =  2  -  4  m.  (2) 

Then,  from  (1)  and  (2),  x  =  20  +  7  m.  (3) 

Any  integral  value  of  m  will  give  to  x  and  y  integral  values. 
But  since  y  is  to  be  positive,  m  <  1  ; 
and,     since  x  is  to  be  positive,  m  >  —  3. 

Therefore  the  only  admissible  values  of  m  are  0,  —  1,  —  2. 
Whenw  =  0,  a  =  20,   y=    2; 

m  =  —  1,          a?  =  13,   ?/  =    6  ; 

m  =  —  2,          0;=    6,   y  =  10. 

3.   An  Indeterminate  System  is  a  system  of  equations  which 
has  an  indefinite  number  of  solutions. 
Thus,  if  the  system        x  -f  y  —  z  =  9, 


be  solved  for  x  and  y,  we  obtain 

x  =  U-2z,y  =  3z-5. 

In  these  values  of  x  and  y  we  may  assign  any  value  to  z 
and  obtain  corresponding  values  of  x  and  y. 

4.  In  solving  a  system  of  two  linear  equations  in  three 
unknown  numbers,  we  first  eliminate  one  of  the  unknown  num- 
bers, and  apply  to  the  resulting  equation  the  preceding  method, 

Pr.  A  party  of  20  people,  consisting  of  men,  womfcn,  and 
children,  pay  a  hotel  bill  of  $  67.  Each  man  pays  $  5,  each 
woman  $  4,  and  each  child  $  1.50.  How  many  of  the  company 
are  men,  how  many  women,  and  how  many  children  ? 

Let  x  stand  for  the  number  of  men,  y  for  the  number  of 
women,  z  for  the  number  of  children. 


212  ALGEBRA.  [Cn.  XII 

Then,  by  the  conditions  of  the  problem, 

aj  +  y  +  «  =  20,  (1) 

6a?  +  4y  +  f3  =  67.    .  (2) 

Eliminating  z,  7  #  +  5  y  =  74. 

Solving  this  equation,  we  obtain 

x  =  2  —  5  m,  y  =  12  +  7  w,  2  =  6  —  2  m. 
When  m  =  0,  x  =  2,  y  =  12,  z  =  6; 

m  =  —  1,       x  =  7,  y=    5,  2  =  8. 

EXERCISES. 

Solve  in  positive  integers  : 

1.  5x+8?/=29.          2.  3a+5y=10.  3.  12a+13y=175. 

4.  25  0^+15  2/=215.    5.  5o,-+132/=229.        6.  34  a;  +  89^=407. 


7z  =  68.          8>  \lx-2y-    z  =  8. 

Solve  in  least  positive  integers  : 
9.  89a;—  144y=l.    10.  14  a;—  49^=133.    11.  67x-43y=5. 

12.  Divide  1000  into  two  parts  so  that  one  part  shall  be  a 
multiple  of  13,  and  the  other  a  multiple  of  53. 

13.  What  positive  integers  when  divided  by  4  give  a  re- 
mainder 3,  and  when  divided  by  5  give  a  remainder  4  ? 

14.  A  farmer  received  $  16  for  a  number  of  turkeys  and 
chickens.     If  he  was  paid  $2  for  each  turkey  and  $.75  for 
each  chicken,  how  many  of  each  did  he  sell  ? 

15.  A  gardener  has  fewer  than  1000  trees.     If  he  plants 
them  in  rows  of  37  each,  he  will  have  8  left  ;  but  if  he  plants 
them  ift  a  different  number  of  rows  of  43  each,  he  will  have  11 
left.     How  many  trees  has  he  ? 

16.  A  said  to  B  :  "  If  I  had  eight  times  as  much  money  as  I 
now  have,  and  you  had  seven  times  as  much  money  as  you 
now  have,  and  I  were  to  give  you  $  1,  we  should  have  equal 
amounts."     How  many  dollars  had  each  ? 


CHAPTER   XIII. 

INVOLUTION 

1.  Involution  is   the   process  of  raising  a  number  to  any 
required  power. 

Powers  of  Powers. 

2.  Ex.  1.    (a4)5  =  a4tt4a4a4a4  =  a4+4+4+4+4  =  a4x5  =  a20. 
Ex.  2.  (a9)10  =  «W  -  -  to  10  factors 

__  ,09+9+9+ -to  lOsummands  _  ^,9x10  __  ^,90 

These  examples  illustrate  the  following  method  of  finding 
any  required  power  of  a  given  power : 

Multiply  the  exponent  of  the  given  power  by  the  exponent  of 
the  required  power ;  or,  stated  symbolically, 

(flm)w  =  amn. 

Eor,  (am)M  =  amamam  •  -  -  to  n  factors 

nm-\-m+m+  —  to  n  summands  flmn 

U/  til          a 

Powers  of  Products. 

3.  Ex.  1.  (aby  =  (a&)  (a6)  (a6)  (a6) 

=  (aaaa)  (bbbb)  =  a4b4. 

Ex.  2   (xy)w  =  (a^)  (xy)  (xy)  -  -  •  to  10  factors 

=  (xxx  ...  to  10  factors)  (yyy  «••  to  10  factors) 
=  xwy™. 

These  examples  illustrate  the  following  method  of  finding 
any  required  power  of  a  product : 

Take  the  product  of  the  factors,  each  raised  to  the  required 
power ;  or,  stated  symbolically, 

(ab)"  =  anbn;    (abc)"  =  a"b"c" ;   etc. 

213 


214  ALGEBRA.  [Cn.  XIII 

For,  (db)»  =  (ab)  (db)  (ab)  -  •  -  to  n  factors 

=  (aaa  •  ••  to  n  factors)  (bbb  -"ton  factors) 
=  anbn. 

In  like  manner,  (abc)n  =  anbncn  ;  and  so  on. 

4.  The  converse  of  the  principle  of  Art.  3  is  evidently  true. 

That  is, 

ambm  =  (ab)m  ;  ambmcm  =  (abc)n  ;  etc. 

5.  The  principles  of  Arts.  2-3  prove  the  method,  already 
given  in  Ch.  V.,  Art.  5,  of  raising  a  monomial  to  any  required 
power. 

Raise  the  numerical  coefficient  to  the  required  power,  and 
multiply  the  exponent  of  each  literal  factor  by  the  exponent  of  the 
required  power. 

Ex.  1.  (4  asb)2  =  42  as*2b2  =  16  «662. 

Ex.  2.       (  -  3  aV)3  =  (  -  3)3  a4xVx3  =  -  27  a%6. 

Powers  of  Fractions. 

F  *\*_2x2     2x>      (2^)2     4z4 

•      -~x    B~SS" 


Ex.  2.  -=  -  x  -  x  -  .-  to  9  factors 

\bj       bbb 

_  aaa  •  •  •  to  9  factors  _  a9 
~bbb  •••to9factors~69' 

These  examples  illustrate  the  following  method  of  raising 
any  fraction  to  a  required  power  : 

liaise  each  term  of  the  fraction  to  the  required  power;  or, 
stated  symbolically, 


For,  -    =  -  x  -  x  -  —  to  w  factors 

\bj       bbb 

_  aaa  •  •  •  to  n  factors  _  an 
~~  bbb  •••  to  n  factors      bn 


3-7]  INVOLUTION.  215 

EXERCISES   I. 

Write  the  cubes  and  the  fourth  powers  of  : 
1.   x2.  2.   —x4.  3.    2x7.  4.    —  Sab. 

5.   5ab2.         6.   4ary.  7-   Zmfaf.  8.    5  aW. 

*§•    -r    »--i£    --is- 

Write  the  squares,  the  cubes,  and  the  ?ith  powers  of  : 

13.   am+\         14.   xm~2.         15.    2xm+ny.         16.    —  3am+n~lif'. 

Find  the  values  of  each  of  the  following  powers  : 

17.    (-3ojV)8.  18.    (5aa68c)2.  19.    (-4xyV)3. 

20.    (2xf#)\  21.    (—cPxf)*.  22.    (-2m2n3)5. 


/3a26V  24  25       _ 

'  '  '  '  ' 


Powers  of  Binomials. 
7.   By  actual  multiplication,  we  obtain 
(a  +  6)3  =  (a2  +  2a6  +  62)  (a  +  b)  =  a3  +  3a26  +  3a62  +  A3, 
(a  _  6)3  =  (a2  -  2ab  +  62)  (a  -  6)  =  a3  -  3a26  +  3a62  -  b3, 
(a  +  6)4  =  (a2  +  2a6  +  62)  (fl2  +  2a6  +  b2) 
=  a4  +  4fl36  +  6a262  +  4a63  +  64, 
(a  -  6)4  =  a4  -  4a36  +  6a262  -  4a63  +  64. 

The  result  of  performing  the  indicated  operation  in  a  power 
of  a  binomial  is  called  the  Expansion  of  that  power  of  the 
binomial. 

In  the  preceding  expansions  the  following  laws  are  evident  : 

(i.)    The  number  of  terms  exceeds  the  binomial  exponent  by  1. 

(ii.)  The  exponent  of  a  in  the  first  term  is  equal  to  the  binomial 
exponent,  and  decreases  by  1  from  term  to  term. 

(iii.)  TJie  exponent  of  b  in  the  second  term  is  1  and  increases 
by  1  from  term  to  term,  and  in  the  last  term  is  equal  to  the 
binomial  exponent. 


216  ALGEBRA.  [Cn.  XIII 

(iv.)  The  coefficient  of  the  first  term  is  1,  and  that  of  the  second 
term,  except  for  sign,  is  equal  to  the  binomial  exponent. 

(v.)  The  coefficient  of  any  term  after  the  second  is  obtained, 
except  for  sign,  by  multiplying  the  coefficient  of  the  preceding  term 
by  the  exponent  of  a  in  that  term,  and  dividing  the  product  by  a 
number  greater  by  1  than  the  exponent  of  b  in  that  term. 

E.g.,  the  coefficient  of  the  fourth  term  in  the  expansion  of 
(a  +  b)4  is  6  x  2  -f-  3,  =  4. 

(vi.)  TJie  signs  of  the  terms  are  all  positive  when  the  terms  of 
the  binomial  are  both  positive  ;  the  signs  of  the  terms  alternate, 
-f  and  —,  ivhen  one  of  the  terms  of  the  binomial  is  negative. 

Observe,  as  a  check  : 

(vii.)  The  sum  of  the  exponents  of  a  and  b  in  any  term  is  equal 
to  the  binomial  exponent. 

(viii.)  The  coefficients  of  tivo  terms  equally  distant  from  the 
beginning  and  the  end  of  the  expansion  are  equal. 

In  a  subsequent  chapter  the  above  laws  will  be  proved  to 
hold  for  any  positive  integral  power  of  the  binomial. 

8.  Ex.  1. 

(2  a  -  3  b)4  =  (2  a)4  -4(2  a)3  (3  b)  +  6(2  a)2  (3  b)2 

_4(2a)(36)3  +  (36)4 
=  16  a4  -  96  a*b  +  216  a262  -  216  ab3  +  81  V. 
Ex.  2. 


EXERCISES  II. 

Raise  each  of  the   following  expressions  to  the  required 
power  : 

1.    (a  +  1)3.  2.    (a-3)3.  3.  (2a  +  3)3. 

4.    (5  -2  y)s.  5.    (2ab  +  3)s.  6.  (5x-6y)3. 

7.    (x2-8)3.  8.    (5^-3i/)3.  9.  (6^ 

10.    (a  -I)4.  11.    (2^  +  3)4.  12.  (3Z-2 

13.    (a  +  b)5.  14.    (2  m  —  3  ?i)^.  15.  (x  —  yf. 


7-9]  INVOLUTION.  217 

Powers  of  Multinomials. 
9.  We  have 

(a  +  b  +  C)2  =  [(a  +  6)  +  c]2  =  (a  +  b)2  +  2  (a  +  6)c  +  c2 

=  .a2  +  2  a&  +  b2  +  2  ac  -f  2  6c  +  c2. 

Therefore       (a  -f  6  +  c)2  =  a2  +  62  +  c2  +  2  a6  +  2  ac  +  2  6c. 
In  like  manner, 

(a  +  b  -  c)2  =  a2  +  62  +  c2  +  2  a6  -  2  ac  -  2  6c. 
(a  _  b  -  c)2  =  a2  +  62  +  c2  -  2  a6  -  2  ac  +  2  fc. 

By  repeated  application  of  this  principle  we  can  obtain  the 
square  of  a  multinomial  of  any  number  of  terms.     We  have 

(a  +  6  +  c  +  d)2  =  [(a  +  b  +  c)2  +  d]2 
=  a2  4.  52  +  C2  +  2  06  +  2  ac  +  2  6c  +  2  (a 
b2  +  c2  +  d* 


That  is,  the  square  of  a  multinomial  is  equal  to  the  sum  of  the 
squares  of  the  terms,  plus  the  algebraic  sum  of  twice  the  product 
of  each  term  by  each  term  which  follows  it. 

Ex.1.  (3x+5y-7zy=(3x)*+(5yy+(-7z)*+2(3x)(oy) 


EXERCISES   III. 

Raise  each  of  the  following  expressions   to  the   required 
power  : 

1.    (a  +  6  +  1)2.  2.    (x-y-T)'2. 

3.    (2a  +  3&  +  l)2.  4.    (3a-4&  +  5c)2. 

5.    (a2  +  a  +  l)*.  6.    02-a  +  l)2. 

7.    (a?  +  xy  +  y*)*.  8.    (a2  -  3  a6  +  &2)2. 

9.    (a  +  6  +  c)3.  10.    (a-6-c)3. 

11.    (a2_a  +  l)3.  12.    (2a-&  +  5)3. 

13.    (a  +  b  +  c  +  d)2.  14.    (a-6-c  +  d)2. 

15.    (a3  -  a2  +  a  -  I)2.  16.    (aj8  +  2x2  -  3^  4-  4)2. 


CHAPTER  XIV. 

EVOLUTION. 

1.  A  Root  of  a  number  is  one  of  the  equal  factors  of  the 
number. 

E.g.,  2  is  a  root  of  4,  of  8,  of  16,  etc. 

2.  A  Second,  or  Square  Root  of  a  number  is  one  of  two  equal 
factors  of  the  number. 

E.g.,  since  5  x  5  =  25  and  (—  5)  (—  5)  =  25,  therefore  +  5 
and  —  5  are  square  roots  of  25. 

A  Third,  or  Cube  Root  of  a  number  is  one  of  three  equal 
factors  of  the  number. 

E.g.,  since  3  x  3  x  3  =  27,  therefore  3  is  a  cube  root  of  27 ; 
since  (—  3)  (—  3)  (—  3)  =  —  27,  therefore  —  3  is  a  cube  root 
of  -  27. 

In  general,  the  qth  root  of  a  number  is  one  of  q  equal  factors 
of  the  number. 

E.g.,  a  qth  root  of  xq  is  x. 

3.  The  Radical  Sign,  -y/,  is  used  to  denote  a  root,  and  is 
placed  before  the  number  whose  root  is  to  be  found. 

The  Radicand  is  the  number  whose  root  is  required. 

The  Index  of  a  root  is  the  number  which  indicates  what  root 
is  to  be  found,  and  is  written  over  the  radical  sign.  The  index 
2  is  usually  omitted. 

E.g.,  ^/9,  or  -y/9,  denotes  a  second,  or  square  root  of  9 ;  the 
radicand  is  9,  and  the  index  is  2. 

4.  A  vinculum  is  often  used  in  connection  with  the  radical 
sign  to  indicate  what  part  of  an  expression  following  the  sign 
is  affected  by  it. 

218 


1-10]  EVOLUTION.  219 

E.g.,  -y/9  4-  16  means  the  sum  of  ^/9  and  16,  while  V9  +  16 
means  a  square  root  of  the  sum  9  +  16.  Likewise  -f/a3  x  bc> 
means  the  product  of  ^/a3  and  66,  while  -\/a3  x  66  means  a  cube 
root  of  a?b6. 

Parentheses  may  be  used  instead  of  the  vinculum  in  connec- 
tion with  the  radical  sign  ;  as  y  (9  4-  16)  for  V9  +  16. 

5.  It  follows  from  the  definition  of  a  root  that  the  square 
of  a  square  root  of  a  number  is  the  number,  the  cube  of  a  cube 
root  of  a  number  is  the  number,  and  so  on. 

E.g.,  (V4)2  =  4;  (-</8)3  =  8;  etc. 
In  general,  (-tya)q  =  a. 

6.  An  Even  Root  is  one  whose  index  is  even;  as  ^/a2,  ^/«4, 

*ya\ 

An  Odd  Root  is  one  whose  index  is  odd;  as  ^/8,  -^/810,  2?^a22+1. 

7.  In  this  chapter  we  shall  consider  only  roots  of  powers 
whose  exponents  are  multiples  of  the  indices  of  the  required 
roots;  as  V^6,  =V42>  ^X> 


Number  of  Roots. 

8.  Since  (±  4)2  =  16,  therefore  V16  =  ±  4  ; 
since  (  ±  a)4  =  a4,   therefore  -^/a4  =  ±  a. 

These  examples  illustrate  the  principle  : 
A  positive  number  has  at  least  two  even  roots,  equal  and  oppo- 
site; i.e.,  one  positive  and  one  negative. 

9.  Since  (-  3)3  =  -  27,  therefore  J/—  27  =  -  3  ; 
since         25  =  32,      therefore       ^/32  =  2. 

These  examples  illustrate  the  principle  : 

A  positive  or  a  negative  number  has  at  least  one  odd  root  of 
the  same  sign  as  the  number  itself. 

10.  Since  (+  4)2  =  -f  16  and  (—  4)2  =  +  16,  there  is  no  num- 
ber, with  which  we  are  as  yet  familiar,  whose  square  is  —  16. 


220  ALGEBRA.  [Cn.  XIV 

Consequently  -^—16  cannot  be  expressed  as  a  positive  or  as 
a  negative  number;  that  is,  in  terms  of  the  numbers  as  yet 
used  in  this  book. 

Such  roots  are  called  Imaginary  Numbers,  and  will  be  con- 

sidered in  Ch.  XVI. 

Evolution. 

11.  Evolution  is  the  process  of  finding  a  root  of  a  given 
number. 

12.  In   the  following  articles   the   radicands  are   limited   to 
positive  values,  and  the  roots  to  positive  roots. 

13.  (i.)  Since  (a2)3  =  a6,  therefore  -fya6  =  a2  =  a*. 
This  example  illustrates  the  principle  : 

The  root  of  a  power  is  obtained  by  dividing  the  exponent  of 
the  power  by  the  index  of  the  root. 

E.g.,  -*/a*  =  a;  -^/a15  =  eft  =  a3. 

5? 

In  general,  Qjanci  =  a  1  =  a". 

nq 

For,  since    (an)q  =  anq,  therefore  J/anq  =  an  =  a'1. 

(ii.)  Since  (a&)2  =  a2b2,  therefore  V(a*^)  =  ab  =  Va'  X  V&2- 
This  example  illustrates  the  principle  : 

The  root  of  a  product  of  two  or  more  factors  is  equal  to  the 
product  of  the  like  roots  of  the  factors,  and  conversely. 

E.g.,         V(16  X  25)  =  V16  X  V25  =  4  X  5  =  20  ; 

,*/(8  a366)  =  -#8  X  s!/a3  x  J/b6  =  2  x  a  x  b2  =  2  ab\ 
In  general,  fy(aib<i)  =  tya*  X  -?/K 

For,  since     (ab)q  =  aqbq,  therefore  -ty(aqbq)  =  ab  — 


(Hi.)  Since    (|)%    ,  therefore         =  |= 

This  example  illustrates  the  principle  : 

The  root  of  a  quotient  of  two  numbers  is  equal  to  the  quotient 
of  the  like  roots  of  the  numbers,  and  conversely. 


10-14]  EVOLUTION.  221 

F  /25  =  ^/25  =  5. 

\16         16     4> 


For,  since      |    =  |,  therefore  $  =  £-£ 


Roots  of  Monomials. 

14.  The  positive  root  of  a  positive  number  can  be  found  by 
applying  the  principles  of  Art.  13. 

The  negative  even  root  of  a  positive  number  is  found  by  pre- 
fixing the  negative  sign  to  its  positive  root. 

Since  ^/-8  =  -2,  and  _^/8  =  -2, 

therefore  ^-8  =  --^/8. 

That  is,  the  negative  odd  root  of  a  negative  number  is  found 
by  prefixing  the  negative  sign  to  the  positive  root  of  the  radi- 
cand  taken  positively. 


Ex.  1.  V(16  a2b4)  =  V16  X  V^2  X 

=  4  ab2,  the  positive  square  root. 

Therefore    ±  V(16  a264)  =  ±  4  ab2. 

In  the  following  examples  we  shall  give  only  the  positive 
even  roots. 

Ex.  2.     j/(-  27  ajtyk8)  =  ^/-  27  x  ^  x  -tyjf  x 


These  examples  illustrate  the  following  method  : 

Tafce  f/ie  required  root  of  the  numerical  coefficient,  and  divide 

the  exponent  of  each  literal  factor  by  the  index  of  the  required 

root. 

Ex  3      4  /16  a8fr12        /16  a8612          /16  a^J^     2  a2 


222  .    ALGEBRA.  [Cn.  XIV 

15.   It  is  frequently  of  advantage  to  separate  a  number  ex- 
pressed in  figures  into  its  prime  factors  before  taking  the  root. 

Ex.  4.    V(15  X  40  x  216)  =  V(o  •  3  x  23  •  5  x  23  •  33) 

=  V(o2  -  34  •  26)  =  5  •  32  •  23  =  360. 

EXERCISES  I. 

Simplify  the  following  expressions : 

3. 
6. 
9. 
12. 


\  cwd2n  '  \    a8616 

Find  the  values  of  each  of  the  following  expressions  : 
25.    V643-  26-    V49"-  27-    A/216'-  28-    S/-274. 

29.    V(40  x  15  x  6).  30.   V(56  x  40  x  35). 

31.  v1024-  32-  V2025-  33-  V12544- 

34.   ^/(6  x  20  x  225).  35.   ^/(84  x  18  x  49). 

36.  V(45  xy  X  35  o?2  x  63  2/2). 

37.  -3/(36  a26c  x  75  a6V  x 


SQUARE   ROOTS  OF   MULTINOMIALS 

16.  The  square  root  of  a  trinomial  which  is  the  square  of  a 
binomial  can  be  found  by  inspection  (Ch.  VI.,  Art.  9). 

17.  Since  (a  +  b)2  =  a2  +  2  ab  +  &2, 
we  have            V(a2  +  2  a6  +  62)  =  a  +  &. 


15-18]          SQUARE  ROOTS  OF  MULTINOMIALS.  223 

From  this  identity  we  infer  : 

(i.)  The  first  term  of  the  root  is  the  square  root  of  the  first 
term  of  the  trinomial;  i.e.,  a=A/a2. 

(ii.)  If  the  square  of  the  first  term  of  the  root  be  subtracted 
from  the  trinomial,  the  remainder  will  be 


Twice  the  first  term  of  the  root,  2  a,  is  called  the  Trial 
Divisor. 

(iii.)   The  second  term  of  the  root  is  obtained  by  dividing  the 

'first  term  of  the  remainder  by  the  trial  divisor  ;  i.e.,  b  =  ~  — 

^  <  ' 

The  trial  divisor  plus  the  second  term  of  the  root  is  called 
the  Complete  Divisor. 

(iv.)  If  the  product  of  the  complete  divisor  by  the  second  term 
of  the  root  be  subtracted  from  the  first  remainder)  the  second 
remainder  will  be  0. 

The  work  may  be  arranged  as  follows  : 


2  ab  +  b2 


2ab 


2ab 


a  +  b 


2  a  trial  divisor 

-  2  a  =  6,  second  term  of  root 
2  a  +  b  complete  divisor 

=  (2a  +  6)6 


18.   Ex.  1.    Find  the  square  root  of  4  a4  -  12  x*y  +  9  f. 
The  work,  arranged  as  above,  writing  only  the  trial  and  the 
complete  divisor,  is : 

4  x4  -  12  x2y  - 


-12x*y 

-  12  o?y  +  9  if 


4  a/*2 


The  square  root  of  4  a?4  is  2  a?2,  the  first  term  of  the  root.    The 
trial  divisor  is  2  (2  a?2),  =  4  x~.     The  second  term,  of  the  root  is 

'•—- -,  =  —  3y.     The  complete  divisor  is  4  x~  —  3  y. 

~r  Ou 


224  ALGEBRA.  [Cn.  XIV 

Ex.  2.   Find  the  square  root  of 

4  x4  -  12  x3  +  29  x2  -  30  x  +  25. 
The  work  follows : 

-  29  x2  -30  a +  25    2x2-3x  +  5 


-12x* 

9x2 


20  x2 

20 x2-  30x4-  25 


Only  the  trial  divisor  and  the  complete  divisor  of  each  stage 
are  written,  the  other  steps  being  performed  mentally. 

The  square  root  of  4  a?4  is  2  x2,  the  first  term  of  the  root.  The 
trial  divisor  is  2  (2  x2),  =  4  x2.  The  second  term  of  the  root  is 

1°  x8 

^— ,  =  —  3  x.     The  complete  divisor  is  4  or2  —  3  #,  which  is 

4  X 

multiplied  by  the  second  term  of  the  root,  giving  — 12  Xs  +  9  x2. 
The  first  term  of  the  second  remainder  is  20  x2. 

20  x2 

The  third  term  of  the  root  is  .  =  5. 

4  a2 

To  form  the  complete  divisor  at  this  stage,  we  multiply  the 
part  of  the  root  previously  found,  2  x2  —  3  x,  by  2,  and  to  the 
product  add  the  term  just  found.  We  thus  obtain  4  x2— 6  cc+5. 
This  complete  divisor  we  multiply  by  the  last  term  of  the  root. 

In  the  preceding  examples  the  terms  were  arranged  to  de- 
scending powers  of  x.  They  could  equally  well  have  been 
arranged  to  ascending  powers. 

19.   The  preceding  method  can  be  extended  to  find  square 
roots  which  are  multinomials  of  any  number  of  terms. 
The  work  consists  of  repetitions  of  the  following  steps : 

After  one  or  more  terms  of  the  root  have  been  found,  obtain 
each  succeeding  term,  by  dividing  the  first  term  of  the  remainder 
at  that  stage  by  twice  the  first  term  of  the  root. 


18-19]  SQUARE   ROOTS   OF   MULTINOMIALS.  225 

Find  the  next  remainder  by  subtracting  from  the  last  remainder 
the  expression  (2  a  +  b)  b,  wherein  a  stands  for  the  part  of  the 
root  already  found,  and  b  for  the  term  last  found. 

EXERCISES   II. 

Find  the  square  root  of  each  of  the  following  expressions  : 
1.    #4-4ar34-8a;  +  4.  2.   4w4-4m3+5m2-2m-f  1. 

3.   x*-2x3  +  3x2-2x  +  l.         4. 
5.   9x4+12or}-26z2-20a;+25.     6.   4x4- 

7.  x4y4  —  4  xsy3  +  6  x2y2  —  4  xy  -f-  1. 

8.  ±x4  +  fafy  +  2a?y-  12  0^  +  90*. 

9.  a;4  -  6  ax3  +  13  oV  -  12  a*x  +  4  a4. 

10.  4  a2  +  9  62  +  16  c2  -  12  a&  +  16  ac  -  24  be. 

11.  49x8H-42iK6-19a;4-12«2  +  4. 

12.  25  a*4  -  30  ax3  +  49  aV  -  24  a;3aj  +  16  a4. 

13.  a4 


2 

14.  9  a4  +  30  o86  +  49  a?W  +  40  a63  +  16  b\ 

15.  89  a262  -  70  ab3  +  16  a4  -  56  a36  +  25  b4. 

16.  4  a6  -  12  a46  -  28  a363  +  9  a262  +  42  ab4  +  49  66. 

17.  ^4_i^ 
2/4       2/ 

O  ^ 

18.  ^  +  ^  + 

a       or 

19.  l  +  2^-x2 

20.  a6  -  6  ax5  +  15  a2x4  -  20  aV  +  15  a4ic2  -  6  adx  + 

21.  1  _  4a  +  64  a6  -  64  a5  -  32  a3  +  48  a4  +  12  a2. 

22.  4  a6  +  17  a2  -  22  a3  +  13  a4  -  24  a  -  4  a5  +  16. 

6  42  2    - 


23.    9x6  +  6  arty  +  43  x4y2  +  2&y  + 
24. 


-2 
x      4  or      or 


226  ALGEBRA.  [Cn.  XIV 

CUBE   ROOTS   OF   MULTINOMIALS. 

20.  The  process  of  finding  the  cube  root  of  a  multinomial  is 
the  inverse  of  the  process  of  cubing  the  multinomial. 

Since  (a  +  b)s  =  a3  +  3  a2b  +  3  ab2  +  W 

=  a5  +  (3  a2  +  3  ab  +  6s)  6,  (1) 

we  have  -^/(a3  +  3  a26  +  3  a&2  +  63)  =  a  +  6.  (2) 

From  the  identity  (2),  we  infer : 

(i.)  The  first  term  of  the  root  is  the  cube  root  of  the  first  term 
of  the  multinomial ;  i.e.,  a=^/a3. 

(ii.)  If  the  cube  of  the  first  term  of  the  root  be  subtracted  from 
the  multinomial,  the  remainder  will  be 

3a2b  +  3ab2  +  b3,  =  (3a2  +  Sab  +  b-)b. 

Three  times  the  square  of  the  first  term  of  the  root,  3  a2,  is 
called  the  Trial  Divisor. 

(iii.)  The  second  term  of  the  root  is  obtained  by  dividing  the 
first  term  of  the  remainder  by  the  trial  divisor ;  i.e.,  b  =  - — -• 

3  CL" 

The  sum  3  a2  +  3  ab  -f  b2  is  called  the  Complete  Divisor. 

(iv.)  If  the  product  of  the  complete  divisor  by  the  second  term 
of  the  root  be  subtracted  from  the  first  remainder,  the  second 
remainder  will  be  0. 

The  work  may  be  arranged  as  follows : 


0*6  +  3  O 


3 


3a2b 


3a2b+3ab2+b3 


a+b 


3  a2  trial  divisor  (1) 

3  a26^-3  a2=&,  second  term  of  root  (2) 

3  a2 + 3  ab  +  62,  complete  divisor  (3) 

b  (4) 


21.   Ex.1.   Find  the  cube  root  of  27  a^+54  x*y+ 36  xy2  f  8/. 
The  work,  arranged  as  above,  is : 

27  x3  +  54  ic2?/  +  36  xy2  +  8y* 

27  y? 

54  x*y 


20-23]  CUBE   ROOTS  OF  MULTINOMIALS.  227 

The  cube  root  of   27  x?  is  3  x,  the  first  term  of  the  root. 
The  trial  divisor  is  3  (3  a-)2  =  27  a2. 

*-    I         9 

The  second  term  of  the  root  is  ^  x^,  =  2  ?/.     The  complete 
divisor  is 


2  +  3(3o?)(2  y)  +  (2y)2,  =  27  x?  +  18  a#  +  4/, 
which  is  multiplied  by  the  second  term  of  the  root,  giving 


22.  The  preceding  method  can  be  extended  to  find  cube 
roots  which  are  multinomials  of  any  number  of  terms,  as  the 
method  of  finding  square  roots  was  extended.     The  work  con- 
sists of  repetitions  of  the  following  steps : 

After  one  or  more  terms  of  the  root  have  been  found,  obtain 
each  succeeding  term  by  dividing  the  first  term  of  the  remainder 
at.  that  stage  by  three  times  the  square  of  the  first  term  of  the  root. 

Find  the  next  remainder  by  subtracting  Irom  the  last  remainder 
the  expression  (3a2  +  3a&  +  bz)b,  wherein  a  stands  for  the  part 
of  the  root  already  found,  and  b  for  the  term  last  Jound. 

23.  The  given  multinomial  should  be  arranged  to  powers  of 
a  letter  of  arrangement. 

Ex. 

27-27z+90z2-55 

27 

-27  x 

-27  x+  9x*- 


81  z2- 


3(3)2+3(3)  (- 


EXERCISES  III. 

Find  the  cube  root  of  each  of  the  following  expressions : 

2.  1  —  6  a; +  12  a2  — So8. 

3.  64  a3  +  240  a~b  +  300  ab2  +  125  b3. 

4.  x6  -  6  x5  +  15  x4  -  20  x*  +  15  y?  -  6  x  +  1. 


228  ALGEBRA.  [Cn.  XIV 

5.  8  x6  -  36  .x5  +  66  x4  -  63  or3  +  33  x2  -  9  x  +  1. 

6.  156  a4  -  144  a5  -  99  a3  +  64  a6  +  39  a2  -  9  a  +  1. 

7.  l  +  3x 

8.  l-6x 


9.   8-12 

8  x3 


10.  27  aV5  +  54  aV  +  9  aV  -  28  aV  _  3  aV  +  6  ax  -  1. 

11.  8  a6  +  48  a5b  +  60  a462  -  80  a363  -  90  a264  +  108  «65  -  27  66. 

12.  x3  +  3  x1  -  9  a11  -  27  x15  -  6  tf  -  54  x13  +  28  x9. 

13.  108  a5  -  48  a4  +  8  a3  +  54  a7  -  12  a8  +  a9  -  112  a6. 

14.  8  a6  -  48  a5x  +  60  aV  -  27  x6  -  108  ax5  -  90  aV  +  80  a3ar». 

15.  i  +  3  x  _  8  x3  -  6  x4  +  6  or5  +  8  x6  -  3  x8  -  x9. 

125  /      150  .y5     165  y4     172  y3     99  y2     54  y     27 
x6  x5  x4  xs  x2    '      x 

ROOTS   OF  ARITHMETICAL  NUMBERS. 
Square  Roots. 

24.  Since  the  squares  of  the  numbers  1,  2,  3,  •••,  9,  10,  are 
1,  4,  9,  ••-,  81,  100,  respectively,  the  square  root  of  an  integer 
of  one  or  two  digits  is  a  number  of  one  digit. 

Since  the  squares  of  the  numbers  10,  11,  •••,  100,  are  100, 
121,  •••,  10000,  the  square  root  of  an  integer  of  three  or  four 
digits  is  a  number  of  tivo  digits  ;  and  so  on. 

Therefore,  to  find  the  number  of  digits  in  the  square  root  of  a 
given  integer,  we  first  mark  off  the  digits  from  right  to  left  hi 
groups  of  two.  The  number  of  digits  in  the  square  root  will  be 
equal  to  the  number  of  groups,  counting  any  one  digit  remaining 
on  the  left  as  a  group. 

25.  The  method  of  finding  square  roots  of  numbers  is  then 
derived  from  the  identity 

(a  +  &)2  =  a2+(2«  +  &)6,  (1) 

wherein  a  denotes  tens  and  b  denotes  units,  if  the  square  root 
is  a  number  of  two  digits. 


23-26] 


ROOTS   OF   ARITHMETICAL   NUMBERS. 


229 


26.    Ex.  1.     Find  the  square  root  of  1296. 

We  see  that  the  root  is  a  number  of  two  digits,  since  the 
given  number  divides  into  two  groups.  The  digit  in  the  tens* 
place  is  3,  the  square  root  of  9,  the  square  next  less  than  12. 
Therefore,  in  the  identity  (1),  a  denotes  3  tens,  or  30. 

The  work  then  proceeds  as  follows  : 


12' 96 
9  00 


3  96 
3  96 


30  +  6  =  36 


2  a  =  60,          trial  divisor 

(2ab  +  62)  -r-  2a  =  396  -=-  60  =  6  + 

=  (2  a  +  b)  x  6  =  (60  +  6)  x  6 


(1) 

(2) 
(3) 


The  first  remainder,  396,  is  equal  to  2  ab  -f  b2,  and  cannot 
be  separated  into  the  sum  of  two  terms,  one  of  which  is  2  ab. 
We  cannot,  therefore,  determine  b  by  dividing  2  ab  by  2  a,  as 
in  finding  square  roots  of  algebraic  expressions.  Consequently 
step  (2)  suggests  the  value  of  6,  but  does  not  definitely  deter- 
mine it.  As  a  rule,  we  take  the  integral  part  of  the  quotient, 
6  in  the  above  example,  and  test  that  value  by  step  (3). 

This  method  may  be  extended  to  find  roots  which  contain 
any  number  of  digits.  At  any  stage  of  the  work  a  stands  for 
the  part  of  the  root  already  found,  and  b  for  the  digit  to  be- 
found. 

Ex.  2.     Find  the  square  root  of  51529. 

The  root  is  a  number  of  three  digits,  since  the  given  number 
divides  into  three  groups.  The  digit  in  the  hundreds'  place  is  2y 
the  square  root  of  4,  the  square  next  less  than  5,  Therefore  in 
the  identity  (1),  a  denotes  2  hundreds,  or  200,  in  the  first  stage- 
of  the  work.  The  work  then  proceeds  as  follows : 


5'  15'  29 
4  00  00 

200  +  20  +  7  =  227 

(1) 
(2) 

(3) 
(4) 

2  a  =  400,          trial  divisor 
(2  ab  +  fc2)  -*-  2  a  =  11529  --  400  =  20  + 
=  (2  a  +  6)  6  =  (400  +  20)  x  20 

1  15  29 

84  00 

31  29 
31  29 

(2  a&  +  &2)  -«-  2  a  =  3129  -s-  440  =  7  + 
=  (2  a  +  6)  5  =  (440  +  7)  x  7 

5'  15'  29 
4 

227 

L  =  2 

11  +  4 
42 

1  15 
84 

31  29 
31  29 

312- 
447 

44  = 

230  ALGEBRA.  [Cn.  XIV 

In  the  second  stage  of  the  work,  a  stands  for  the  part  of  the 
root  already  found,  220,  and  b  for  the  next  figure  of  the  root. 
In  practice  the  work  may  be  arranged  more  compactly,  omitting 
unnecessary  ciphers,  and  in  each  remainder  writing  only  the 
next  group  of  figures.  Thus  : 


(2) 
(4) 


Observe  that  the  trial  divisor  at  any  stage  is  twice  the  part  of 
the  root  already  found,  as  in  (2)  and  (4). 

27.  The  abbreviated  work  in  the  last  example  illustrates 
the  following  method : 

After  one  or  more  figures  of  the  root  have  been  found,  obtain 
the  next  figure  of  the  root  by  dividing  the  remainder  at  that  stage 
(omitting  the  last  figure),  by  the  trial  divisor  at  that  stage. 

See  lines  (2)  and  (4). 

Annex  this  quotient  to  the  part  of  the  root  already  found,  and 
also  to  the  trial  divisor  to  form  the  complete  divisor. 

Find  the  next  remainder  by  subtracting  from  the  last  remainder 
the  product  of  the  complete  divisor  and  the  figure  of  the  root 
last  found. 

28.  Since  the  number  of  decimal  places  in  the  square  of  a 
decimal  fraction  is  twice  the  number  of  decimal  places  in  the 
fraction,  the  number  of  decimal  places  in  the  square  root  of  a 
decimal  fraction  is  one-half  the  number  of  decimal  places  in 
the  fraction. 

Consequently,  in  finding  the  square  root  of  a  decimal  frac- 
tion, the  decimal  places  are  divided  into  groups  of  two  from 
the  decimal  point  to  the  right,  and  the  integral  places  from 
the  decimal  point  to  the  left  as  before. 


26-30]        ROOTS  OF   ARITHMETICAL  NUMBERS.  231 

Ex. 


14'  46.28'  09 
9 

38.03 
68 

5  46 
5  44 

2.28  09 
2.28  09 

76.03 

In  finding  the  second  figure  of  the  root,  we  have  -5^  =  9 ;  but 
69  x  9  =  621,  which  is  greater  than  546,  from  which  it  is  to  be 
subtracted.  Hence  we  take  the  next  less  figure  8. 

EXERCISES  IV. 

Find  the  square  root  of  each  of  the  following  numbers : 
1.   196.          2.   841.          3.   1296.          4.   65.61.          5.   7396. 
6.   3481.  7.   667489.  8.   170569.  9.   1664.64. 

10.   582169.      11.   1.737124.       12.   556.0164.       13.    .00099225. 

Cube  Roots. 

29.  Since  the  cubes  of  the  numbers  1,  2,  3,  •••,  9,  10,  are  1, 
8,  27,  •••,  729,  1000,  respectively,  the  cube  root  of  any  integer 
of  one,  two,  or  three  digits  is  a  number  of  one  digit.      The  cube 
roots  of  such  numbers  can  be  found  only  by  inspection. 

Since  the  cubes  of  10,  11,  •••,  100  are  1000, 1331,  •••,  1000000, 
respectively,  the  cube  root  of  any  integer  of  four,  Jive,  or  six 
digits  is  a  number  of  two  digits,  and  so  on. 

Therefore,  to  find  the  number  of  digits  in  the  cube  root  of  a 
given  integer,  we  first  mark  off  the  digits  from  right  to  left  in 
groups  of  three.  The  number  of  digits  in  the  cube  root  will 
be  equal  to  the  number  of  groups,  counting  one  or  two  digits 
remaining  on  the  left  as  a  group. 

30.  The  method  of  finding  cube  roots  of  numbers  is  derived 
from  the  identity 

(a  +  b)3  =  a3  +  (3  a2  +  3  ab  +  b2)  b,  (1) 


232 


ALGEBRA. 


[On.  XIV 


wherein  a  denotes  tens,  and  b  denotes  units,  if  the  cube  root 
is  a  number  of  two  digits. 

Ex.   Find  the  cube  root  of  59319. 

The  digits  in  the  tens1  place  of  the  root  is  3,  the  cube  root 
of  27,  the  cube  next  less  than  59.  Therefore  in  identity  (1), 
a  denotes  3  tens,  or  30.  The  work  may  be  arranged  as  follows : 


59'319 
27000 

a  +  b 
30  +  9 

(1) 
(2) 

(3) 

32  319 
32319 

3a2  =  3(30)2  =  2700 
(3  a2b  +  3  ab2  +  b3)  -  3  a2  =  32319  +  2700  =  9  + 
3a2  =  3(30)2    =2700 

b-  =           92  =      81 

=  (3  a2  +  3  ab  +  b2)  xb=  3591  x  9 

As  in  finding  square  roots  of  numbers,  step  (2)  suggests  the 
yalue  of  b,  but  does  not  definitely  determine  it.  If  the  value 
of  b  makes  (3  a2  +  3  ab  +  b2)  x  b  greater  than  the  number  from 
which  it  is  to  be  subtracted,  we  must  try  the  next  less  number. 

In  practice  the  work  may  be  arranged  more  compactly, 
omitting  unnecessary  ciphers,  and  in  each  remainder  writing 
only  the  next  group  of  figures ;  thus 


(1) 

(2) 
(3) 


31.  The  preceding  method  may  be  extended  to  find  roots 
that  contain  any  number  of  digits. 

At  any  stage  of  the  work  a  stands  for  the  part  of  the  root 
already  found,  and  b  for  the  digit  to  be  found. 

The  method  consists  of  repetitions  of  the  f  ollowing^  steps  : 

The  trial  divisor  at  any  stage  is  three  times  the  square  of  the 
part  of  the  root  already  found  ;  as  27  in  the  preceding  example. 


59'  319 

27 

39 

2700 
810 
81 

32  319 
32  319 

3591 

30-33]  HIGHER   ROOTS.  233 

After  one  or  more  figures  of  the  root  have  been  found,  obtain 
the  next  figure  of  the  root  by  dividing  the  remainder  at  that  stage 
{omitting  the  last  two  figures)  by  the  trial  divisor.  In  the  last 
example,  9  +  =  323  -r-  27. 

Annex  this  quotient  to  the  part  of  the  root  already  found. 

To  obtain  the  complete  divisor,  add  to  the  trial  divisor  (with  two 
ciphers  annexed)  three  times  the  product  of  the  part  of  the  root 
already  found  (with  one  cipher  annexed)  by  the  figure  of  the  root 
just  found,  and  also  the  square  of  the  figure  of  the  root  just 
found. 

Find  the  next  remainder  by  subtracting  from  the  last  remainder 
the  product  of  the  complete  divisor  and  the  figure  of  the  root  last 
found. 

32.  Evidently,  in  finding  the  cube  root  of  a  decimal  fraction 
the  decimal  places  are  divided  into  groups  of  three  figures  from 
the  decimal  point  to  the  right,  and  the  integral  places  from  the 
decimal  point  to  the  left  as  before. 

EXERCISES  v. 

Find  the  cube  root  of  each  of  the  following  numbers : 
1.  2744.          2.  39304.  3.  110.592.          4.  328509. 

5.  1.191016.   6.  74088000.     7.  340068392.     8.  426.957777. 
9.  584067.412279.     10.  375601280.458951.   11.  .041063625. 

HIGHER   ROOTS. 

33.  Since      -yVa4  =  a,  and 
therefore,  -yVa4  =  - 

Since  -«/aG  =  a,  and  -fy^/ a*  =  $/<$  =  a, 

therefore,  Va6 

In  general,  since 

tyaM  =  a,  and 
therefore,  **/a?q 


234  ALGEBRA.  [Cn.  XIV 

That  is,  the  pqth  root  of  a  number  is  the  pth  root  of  the  qth 
root  of  the  number. 

In  particular,  the  fourth  root  is  the  square  root  of  the  square 
root,  the  sixth  root  is  the  cube  root  of  the  square  root. 

EXERCISES  VI. 

Find  the  fourth  root  of  each  of  the  following  expressions  : 


2. 

+  16  asb5  +  10  a2b6  +  4  ab7  +  b8. 

3.  16  a8  -  160  a7  +  408  x6  +  440  or5-  2111  x4 

-  1320  or3  +  3672  x2  +  4320  x  +  1296. 

4.  625  x8  +  5500  x7  +  17150  x«  +  20020  aj5 

+  721  x4  -  8008  x3  +  2744  x2  -  352  x  +  16. 

Find  the  sixth  roots  of  each  of  the  following  expressions  : 

5.  64  x12  -  192  x10  +  240  x8  -  160  x6  +  60  x4  -  12  x2  +  1. 

6.  a12+6  ou6+21  a1062+50  a9 

+126  a*67+90  a468+50  o8 


Find  the  value  of  each  of  the  following  indicated  roots  : 

7.    ^7279841.        8.    ^3010936384.         9.    ,J/164204746.7776. 


CHAPTER   XV. 

SURDS. 

1.  In  Ch.  XIV.  we  considered  only  roots  of  powers  whose 
exponents  were  multiples  of  the  indices  of  the  required  roots. 
Such  roots  as  -y/2,  -y/a2,  etc.,  were  excluded. 

2.  It  is  proved  in  School  Algebra,  Ch.  XVIII.,  that  -y/2, 
-^/a2,  etc.,  cannot  be  expressed  either  as  integers  or  as  fractions. 
Thus,  there  is  no  integer  or  fraction  whose  square  is  2. 

But  it  is  there  proved  that  the  value  of  such  a  root  can  be 
found  approximately  to  any  degree  of  accuracy. 

E.g.,  approximate  values  of  ^/2  are  1.4,  1.41,  1.414,  etc. 

3.  It  is  also  proved  that  these  roots  obey  the  fundamental 
laws  of  Algebra  ;  as  ^/2  x  ^/3  =  ^/3  x  -^2,  etc. 

4.  An  Irrational  Number  is  a  number  which  cannot  be  ex- 
pressed as  an  integer  or  as  a  fraction  ;  as  ->/2,  -^/a2. 

An  Irrational  Expression  is  an  expression  which  involves  an 
irrational  number  ;  as  -y/o,  a 


5.  A  Rational  Number  is  a  number  which  can  be  expressed  as 

2  x 

an  integer  or  as  a  fraction;  as  2,  -  —  ,  ^/(27a6). 

3y 

A  Rational  Expression  is  an  expression  which  involves  only 
rational  numbers  ;  as  -|  a  +  ^  b,  ab  -f-  ->/a2. 

6.  A  Radical  is  an  indicated  root  of  a  number  or  expression  ; 

as  y'7,  V9>  ^(a  +  6). 

A  Radical  Expression  is  an  expression  which  contains  radi- 
cals; as  2V?,  V^+V^  V(a  +  V&)- 

7.  A  Surd  is  an  irrational  root  of  a  rational  number;  as 


236  ALGEBRA.  [Cn.  XV 


Observe  that  -^/(l  +  y'T)  is  not  a  surd,  since  1  -f-  -y"7  is  not  a 
rational  number. 

8.  The  Order  of  a  surd  is  indicated  by  the  index.    Thus,  y'a 
is  surd  of  the  second  order,  or  a  quadratic  surd;  -^/5  is  a  surd 
of  the  third  order;  and  so  on. 

Principles  of  Surds. 

9.  As  in  Ch.  XIV.,  we  limit  the  radicands  to  positive  values, 
and  the  roots  to  positive  roots. 

10.  The  principles  established  in  Ch.  XIV.,  Art.  13,  and 
their  proofs,  hold  also  for  surds.  For,  any  positive  number  is 
a  power  of  either  a  rational  or  an  irrational  number. 

Thus,  4  =  22,  3  =  (  V3)2,  a  =  (-tya)*. 

We  have  ^(ab)  =  VC(Va)2(V6)2]  =  Va  X  V&  5  and  so  on- 
Therefore, 

(i.)          ^/a"<?  =  tf  =  a".         [Ch.  XIV.,  Art.  13,  (i.).] 
(ii.)      ty(ab)  =  tya  x  tyb.     [Ch.  XIV.,  Art.  13,  (ii.).] 

.  XIV.,  Art.  13,  (iii.).] 


Reduction  of  Surds. 

11.  A  surd  is  in  its  simplest  form  when  the  radicand  is  in- 
tegral, and  does  not  contain  a  factor  with  an  exponent  equal  to 
or  a  multiple  of  the  index  of  the  root;  as  -^/2,  -J/(cr6),  -\/am. 


12.  Ex.  i.    V80  =  V(16  x  5)  =  V1(5  x  V5  =  4  V5- 

Ex.  2.    V(18  a^2)  =  V(9  a^2  X  2  a)  =  V(9  «4&2)  X  V(2  a) 

=  3  a2b  V(2  a). 

These  examples  illustrate  the  following  method  of  reducing 
a  surd  to  its  simplest  form  : 

Separate  the  radicand  into  two  factors,  one  a  product  of  powers 
with  the  highest  exponents  which  are  multiples  of  the  given  index. 
Multiply  the  rational  root  of  this  factor  by  the  irrational  root  of 
the  second  factor. 


7-13]  SURDS.  237 

Ex.  3.    sX48  xY)  =  J/(S  ofy8  x  6  x2)  = 
Ex.  4.    ^(an+162n+2)  =  y(anb-n  x  a&2)  =  a 

EXERCISES   I. 

Eeduce  each  of  the  following  surds  to  its  simplest  form  : 

i.  v32-         2-  V75-         3-  V108-  4- 


5.  VW-        6-    V(a4&5)-       7 

9.  ^/192.        10.    -^-101     11.    -4/-a10.        12. 

13.  sX16aV).  14.    -^/(32aV).  15. 

16.  V^"+4-  17-    n+\/a2n+3-  18-    " 

19.  V(a2n&2M+1)-  2a    ^/(-»7n^3n.  21. 

22.  a2&2  +  a*<?-  23' 


24.    V(6  -  c)  (W  -  c3).  25.    V(a2  -!)(!  +  a). 

26. 


13.  When  the  Expression  under  the  Radical  Sign  is  a  Frac- 
tion. —  In  this  case  we  reduce  the  numerator  and  denominator 
separately  by  Art.  10  (Hi.). 

13  a2 


8^_     /8a^_ 

-  —  -\/  —  ^—  — 

\  / 


When  the  required  root  of  the  denominator  is  not  rational, 
we  proceed  as  in  Ex.  2  : 

First  multiply  both  terms  of  the  fraction  by  the  expression  of 
lowest  degree  tvhich  will  make  the  denominator  a  power  with  an 
exponent  equal  to  the  index  of  the  root.  Then  proceed  as  before. 

Ex  3     3/7-3/7      _»/7x2x9_^/126 

=*          - 


EXERCISES   II. 

Simplify  each  of  the  following  expressions : 

15  las  3/32orV  4/ 

IB  \9     2-  \r       Viz?'       V 


238  ALGEBRA.  [Cn.  XV 


5. 

Vi- 

6. 

2Vi- 

7. 

Vi- 

8. 

6Vf- 

9. 

VI- 

10. 

Vf 

11. 

6Vf 

12. 

8VI- 

13. 

vt- 

14. 

Vf 

15. 

Vf 

16. 

Vsla- 

17. 

/64ct 

18. 

118  aV 

19. 

/16a8 

20. 

V?' 

\  125  b5 

\f456V 

21. 

3  la 

w 

22. 

3  /  a 

23. 

3/3aV 

24. 

3  /0^4n_ 

VWh 

\4  #y 

\862 

25. 

3/128aV 

26. 

4/16aV6 

27. 

5  /a668 

28. 

6/'      a« 

\    W8 

\   by1 

\  a;8  ' 

14.   When  the  index  of  the  root  and  the  exponent  of  the  radi- 
cand  have  a  common  factor.     We  have 

(^/a2)12  =[(^/a2)3]4  =[a2]4  =  a8. 

Therefore,  ^/a8  =  ^a2  =  ^o*. 

This  example  illustrates  the  following  method  : 
Divide  the  index  of  the  root  and  the  exponent  of  the  radicand 
by  their  H.  C.  F. 


In  general, 

For, 

Therefore,  ^/anp  =  ^/ap. 

Ex.1.    -i/a2=V«- 

Ex.  2.    j/9  =  ^32  =  4/3. 

Ex.  3.    -^/(27  a3?)6)  =  ^66  x  ^(3  a)3  =  b  V(3  a). 


EXERCISES   III. 

Simplify  each  of  the  following  expressions  : 
1.    ^/25.  2.    -4/49.  3.    ^/8.  4. 

5.    4/16.  6.    4/81.  7.    4/(81a2).  8.    4/(27  a3). 

9.    4/(4aV).  10.    4/'(125aV).  11. 

12.    4/(8a9615).  13.    ^/(64aV°).  14. 

16   */  32  -  17 

16< 


13-16]  SURDS.  239 

Addition  and  Subtraction  of  Surds. 

15.  Similar  or  Like  Surds  are  rational  multiples  of  one  and 
the  same  simple  monomial  surd  ;  as  -^12,  =  2-^/3,  and  5-^/3. 

The  rational  factor  is  called  the  coefficient  of  the  surd  factor. 

16.  Like  surds,  or  such  surds  as  can  be  reduced  to  like  surds, 
can  be  united  by  algebraic  addition  into  a  single  like  surd. 

Ex.  i.  v12+2V27-9V48=2V3+6V3-36V3=-28V3- 

Ex.  2.   8^40  4-  3^135-  2^/625  =  16^5  4-9^5-10^5 


Ex.  3.  v2  -  VI  +  V-°2  =  V2  -  i  V2  +  TW2  =  IV2- 
Ex.  4. 


These  examples  illustrate  the  method  :  Reduce  each  surd  to 
its  simplest  form,  and  take  the  algebraic  sum  of  the  coefficients. 


EXERCISES   IV. 
Simplify  : 


l.   5  v2  +  3  V2  -  7  V2-  2- 

3.   8^9-3^/9  +  7-^9.  4.   2^-5^- 

s.  V5+V20-  6-  V9°-5V40-     7- 

8.    8V(96)-3V(16&)-    9-    </16-3^54.    10.   2^/81-5^/24. 
11.    2^/(8a?)+5-^(3aj).  12.   6  ^/  (108  a)  -  3  y  (500  a). 

is.  x^(xf)+y^/(y?y).  14.  5«V(3^2)-  &V(48«2)- 

is.  v2  +  3V8-V50-  16-  3y3+V27-1:LV48- 

17.   3V64- 
19.   3V754-4W192-2fV12.     20. 

21.  4v|-fVA-2V2^         22.  2 

23.   8^/48  4-  3^/162  -  2^384.      24.    5-^/544-9^250-^/686. 
25.   2f  ^/500  +  f  -4/256  -  31^32  -  f-^ 
26. 


240  ALGEBRA.  [Cn.  XV 

27.   2  V3  -  V12  +  A/9-  (/  -\/24  +  3-^9  -  5,5/192. 


29.  V(4  «3)  +  V(9  «*)  +  V(25  a8)  -  V(81  a3)- 

30.  V(12  «2&)  +  V(75  a2&)  -  V(27  a2&)- 

31.  -^(64  a865)  +  iV(125  a865)  -  ^/(a865). 

32.  a  V(a3^7)  4- 

33.  ^/(9  a«62)  +  V(27  a86)  +  5  ^/(729  a662). 

34.  V(9  «  +  27)  +  3  V(4  a  +  12). 

35.  4  a'3  +  4  a26   +       4  a&2  +  4  63. 
36. 

37. 


Reduction  of  Surds  of  Different  Orders  to  Equivalent  Surds  of 
the  Same  Order. 

17.   The    converse   of   the   principle   of  Art.   14   evidently 
holds.     That  is, 


Ex.  1.    Reduce  ^/2,  -j/3  a,  and  ^/(o  b)  to  equivalent  surds  of 
the  same  order. 

We  have  V2  =  ^       =  ^v 


We  thus  have  the  following  method  : 

Take  the  L.  C.  M.  of  the  given  indices  as  the  common  index 
of  the  equivalent  surds.  Raise  each  radicand  to  a  power  whose 
exponent  is  equal  to  the  quotient  obtained  by  dividing  this  L.  C.  M. 
by  the  given  index. 

18.  Any  rational  number  can  be  expressed  in  the  form 
of  a  surd. 


Ex.2.    2=V4=^3/8  =  -";  a=Va2  =  -v/a3  = 
Write  under  the  radical  sign  a  power  of  the  number  whose 
exponent  is  equal  to  the  index. 


16-20]  SURDS.  241 

19.  Two  surds,  or  a  surd  and  a  rational  number,  can  be 
compared  by  first  reducing  them  to  equivalent  surds  of  the 
same  order. 

Ex.  3.    Which  is  greater,  y2  or  v^  ? 

We  have  y2  =^8,  and  j/3  =  -^/9. 

Since  9  >  8,  therefore  ^/9>^/8,  or  -^/3>V2- 

EXERCISES  V. 

Reduce  to  equivalent  surds  of  the  same  order: 
l.  V2,  </5.  2.  V3,  ^/6.  3.  y7,  -v/10. 

4-  Vi>  A/i  5-  5>  -v/10-  6-  6      /4. 

7.  -«/2        3.  8.     /15    «10.  9. 

10.  v 

Which  is  the  greater, 

13.  2V3  or  3V2?       14.  V5  or  A/IO?     15-  iv"25  or  iV11? 
16.  ^/'a2  or  y  a,  when  a  <  1  ?        17.  -^Ac3  or  ^/x4,  when  x  >  1  ? 

W^hich  is  the  greatest, 
18.  V3>  -v/5'  or  ^1()?  19-  Vf>  ^/i  or  A/P 


Multiplication  of  Surds. 

20.  Multiplication  of  Monomial  Surds.  —  The  converse  of  the 
principle  of  Art.  14  evidently  holds.  That  is, 

^/ax^/b=^/(ab). 
Ex.  1.   5^/4  x  2^/6  =  10^/24  =  20^/3. 

Ex.  2.    Va  x  ->X  -  -\/a3  X  ^a4  =  ^  =  a^a' 

We  thus  have  the  following  method  : 

Reduce  surds  of  different  orders  to  equivalent  surds  of  the  same 
order. 

Multiply  the  product  of  the  coefficients  by  the  product  of  the  surd 
factors. 

Simplify  the  result. 


242  ALGEBRA.  [On.  XV 


Ex.  3.    V12  X  v36  =  V(4  X  3)  x  s^(4  x  9)  =  2  V3  x  ^(22  x  32) 
=  2^/33  x  -4/(2<  x  34)  =  2^/(24  x  37) 
=  6^/(24x  3)  =  6-4/48. 


When  the  radicands  contain  numerical  factors  it  is  advisable 
to  express  them  as  powers  of  the  smallest  possible  bases. 

21.   It  is  frequently  desirable  to  introduce  the  coefficient  of 
a  surd  under  the  radical  sign. 


Ex.  4.    4  Vo  =  V16  x  V5  =  V80- 
Ex.  5.   3  a-ty(2  db)  =  ^(27  a3)  x 


EXERCISES   VI. 

Multiply  : 


4.  4V15XV45.        5.  Vt 

7.  3^/45x5^/150.  8.   9^/54x3^/24. 

9.  ^/6  x  3^/36.        10.  Va  x  V(2  a)-      u-   5  Vm  x 

12.  7V(6^)x4V(lBic).  13.    -^(a2x)x/a. 

14.  ^(5a?»)x-4/(25ajy).  15.    ^/(4  a26)  x 

16.  V(1+^)xV(^  +  a)-         17-    ^/C1  ~  x?  x 

18.  V6  X  -\/4'  19'  A/50  X  A/75'          20'     V21  X  A/27 

21.  -^/20  X  V2'  22'  A/72  X  ^108.       23.    -^'2  X  V3- 


27.    -^/54  x  3  V6  X  5-4/2.  28.   V10  X  -4/100  x  </500. 

29.    ^/12  x  -4/108  x  -4/486.         30.   12^/14  x  V2T  X 

31.  -»/12  x  </216  x  -^96. 

32.  V(40  »)  x  -4/(250  x)  x 


20-23]  SURDS.                                          243 

In  each  of  the  following  expressions  introduce  the  coeffi 
cient  under  the  radical  sign  : 

33.   3V2.  34.   5V3.             35.   2^/25.            36.   10^/7. 

37.    5-^3.  38.    iV2-               39-    *^4-                 40.    iv¥* 

41.   2a^/a.  42.   5ajV(3«y).         43.   4 

44.    ay  a.  45.    a2bn-J/(ab).          46.    a 


/     ,  ,  x  ab  f  N    /m  +  n 

47.      a  +  &—  —  -         48.      m-n 


22.   Involution  of  Monomial  Surds.  —  We  have 

(^/a)3  =  ^/a  x  Va  x  Va  =  V(aaa)  =  Va3- 
In  general,        (-ya)n  =  ^/a  x  -{/a  X  -(/a  •••  to  n  factors 

=  -^/(aaa  •••  to  n  factors) 
or,  (^/a)«=>r». 

That  is,  to  raise  a  surd  to  any  required  power  : 
Raise  the  radkand  to  the  required  power. 

Ex.  6.      -/24 


23.   If  the  index  of  the  root  and  exponent  of  the  required 
power  have  a  common  factor,  the  work  is  simplified  by  Art.  14  : 


Ex.1.    (^/5)2=-^/5.        Ex.2. 
Ex.  3.    [5  x 


EXERCISES   VII. 

Simplify  : 

1.    (V5)2.  2.  (s»3.  3.  (A/^)2.  4.    (-v/^) 

5.    (-\/2^)6.  6.  (V3x)3.  7.  (-v/Sa)2.  8.   (3Va«) 

9.     22.  10.     A/aV2.  11.  3/25.  12. 


13.    (Va46)2.       14.    (VSarV)3.       15.    (V7a)3.      16.    (2aV36) 


244  ALGEBRA.  [Cn.  XV 

24.  Multiplication  of  Multinomial  Surd  Numbers.  —  The  work 
may  be  arranged  as  in  multiplication  of  rational  multinomials. 

Ex.   Multiply  2  V5  +  3  V2  by  V5  —  4V2- 
We  have  2^/5  +  3^/2 

V5  -  4  V2 
10  +  3  V10 
_  8  ylO  _  24 

10  -  5  V10  -  24  =  -  14  -  5  V10. 

25.  Conjugate  Surds.  —  Two  binomial  quadratic  surds  which 
differ  only  in  the  sign  of  a  surd  term  are  called  Conjugate  Surds. 

E.g.,  V3  +  V2  and  ~  V3  +  V25  1  -V5  and  *  +V5- 
Either  of  two  conjugate  surds  is  the  conjugate  of  the  other. 
The  product  of  two  conjugate  surds  is  a  rational  number. 
For,     ( 


26.   Type-Forms.  —  Many  products  are  more  easily  obtained 
by  using  the  type-forms  given  in  Ch.  V. 

Ex.       (v2  +  V3)2  =  (V2)2  +  2  V2  x  V3  +  (V3)2 


EXERCISES  VIII. 

'Simplify  each  of  the  following  expressions  : 

1.  (v3  +  3V6  -  5V8)  x  V6- 

2.  (V9  -  2  V45  +  5  V54)  x  V3- 

;3.  (5+v3)  (l-3V3)-  4-  (V10  -  2)  (V10  +  5)- 

.5.    (2  V7  -  5yi3)  (V91  -  5).    6.    (V6+HV5)(V2+4V15)- 


9. 
jo. 


24-26]  SURDS.  245 


11. 

12.  (  V7 

13. 

14. 

15.  (5^/9  +  3^25)  (-t/3  - 

16.  GJ/27-  -^2)  (2^/3 

Find  the  value  of  each  of  the  following  expressions,  without 
performing  the  actual  multiplications  : 
17-    (V5-V10)2-      18.    (y6-4^/40)2.     19. 
20.    (V6-2^/2)3.      21.    (1  +  V2-V3)2.    22. 
23. 

24.     25- 
25. 

26.   ^/(2  V2  -  3)  x  ^(2  V2  4-  3). 
27-    [  V(7  +  2  V1(>)  -  V(7  -  2V10)]2- 
28.    [  Va  +  V(«2  -  ^)  +  Va  - 
29. 
30. 


31-    V(5  +  V7)  x  V(2  -  V2)  x  V(5  -  V7)  x  V(2  +  V2)- 
32. 


33. 

Simplify  each  of  the  following  expressions : 

id.       t(rt*      M\  v     /**  "*" u  IK       /(&  rj-      K\  v     /^» «/  —  3 

«s*.    \r  ^u  —     /  x  -\  / *  &^'    ~\/  (y  •»  ~  '  v)  x 


36. — X       " 


/y»-<  / 

~^2l-\l 


2  xz  +  4  z2 


„.    ,,  +  f+%lf  -gV.  +  E-^-,). 


246  ALGEBRA.  .  [Cii.  XV 


Division  of  Surds. 


27.   Division  of  Monomial  Surds.  —  The  converse  of  the  prin- 
ciple of  Art.  10  (iii.)  evidently  holds.     That  is, 


2       3^a2  3->4     _  3  6/  a4  _  3  6/33  •  a  _  ,  6/f97  „, 

'   43  a  ~  4-/33a3  ~  4\3V  ~  4\~3^"  = 


) 

We  thus  have  the  following  method  : 

Reduce  surds  of  different  orders  to  equivalent  surds  of  the  same 
order. 

Multiply  the  quotient  of  the  coefficients  by  the  quotient  of  the 
surd  factors. 

Simplify  the  result. 

EXERCISES   IX. 

Simplify  each  of  the  following  expressions  : 

i.  v60-V5-  2.  v15-Vt- 

3-  V-V-Vf  4- 

5.  V(45^HV(5*)-  6- 

7.  ^/x-*-tyx.  8.  V^-*-^2-  9- 

10.    V30^-^/45-  1:L-   3V5^-^/15.          12. 

13.   6V2^-^/9.  14.    2-^/6^-^/2.  15. 

16.    V(14^)^-^(28a2&2)'  17-    ^/(15^)^-^/(25a:?/2). 

18.    (V6-5V14)-V2-  19-    (3V10-4V15HV5- 

20.    (V6-3^4)-j-^2.  21. 

22.  (3  V20  +  2  V15  -  4  V5)-  V10- 

23.  (6-^4  -8^36  -15-^/48)  - 
24.    y(6»-a2)-i-V(a  +  6)-  25-    ^(a26 
26.    x^+2^-a?2-         27-     ^2- 


27-31]  SURDS.  247 

28.   To  Rationalize  a  surd  expression  is  to  free  it  from  irra- 
tional numbers. 

Thus,  -J/4  is  rationalized  by  multiplying  it  by  -J/2,  since 


29.   The  quotient  of  one  surd  divided  by  another,  expressed 
as  a  fraction,  may  be  simplified  by  rationalizing  its  denominator. 

Fv   l     V5  -  V5  X  V3  -  V15  - 
EX<1>    V3~V3^V3"     3     " 

We  thus  have  the  following  method  : 

Multiply  the  numerator  and  denominator  by  a  factor  which  will 
rationalize  the  denominator. 

Ex  2      2Va          2yaxs:/(2a)    =  2-^/a3  x  #(4  a2) 
a2)  x  ^(2  a)  ^  (8  a3) 


30.   The  Divisor  a  Binomial  Quadratic  Surd.  —  We  express  the 
quotient  as  a  fraction  and  rationalize  the  denominator. 
Ex.1. 


5  V2  +  4  V3      (5  V2  +  4  V3)  (5  V^  -  4 

30-2y6-24       6-2y6  =  3         f 

=        "V 


50-48 

We  thus  have  the  following  method  : 

Multiply  the  numerator  and  denominator  by  the  conjugate  of 
the  denominator. 
Ex.  2. 


(l+aj)_(l_aj)  a; 

31.   When  the  denominator  contains  three  quadratic  surds, 
a  similar  method  may  be  employed.  . 


248  ALGEBRA.  [Cu.  XV 

Ex.  3. 

V2  V2(2V3-V2 


2V3-V2+V5    [(2V3- 

_  2y6-2-yio 


12-4V6+2-5  "        9-4V6 

(2ye-2-  yiO)(9+4y6) 


EXERCISES   X. 

Change  each  of  the  following  fractions  into  an  equivalent 
fraction  with  a  rational  denominator : 

12  8  10 


V2               '    V3 
5      x  .             6        ax 

3^/4 
7 

'   7^/25 
a 

'           ,x                          '        3/(aV 

9.    :  10. 

i-    3        (Q 

L                             4  /  '/  /vT$ 

12 

3\                               n    /    n—''    °,\ 
11                  5 

5+V21 

2-V2* 
3V5-2V2 

'    4+VH 

•    5-2V6 
15    5V2-4V3 

5  V3  -  3  V? 

'    5V2  +  4V3 
10                            1 

2V5  -  3V2 

19 

+  2V(2a-l) 

°o 

21 

3  +  4V3 

/1  n        /'/*        /^ 

-y/-L\J    V          ^"yt-' 

V6+V2-V5 

Surd  Factors. 
32.   The  expression 

x2  +  2  ax  +  a2 

is  evidently  the  square  of  x  -f  a.     The  third  term  of  this  ex- 
pression may  be  obtained  as  follows  : 


31-33]  SURDS.  249 

That  is,  the  third  term  is  the  square  of  half  the  coefficient  of  x. 

Consequently,  if  to  any  binomial  of  the  form  x2  +  2  ax,  we 
add  the  square  of  half  the  coefficient  of  a?,  the  resulting  tri- 
nomial will  be  the  square  of  a  binomial. 

This  step  is  called  completing  the  square. 

Thus,  if  to  x2  +  6  a?,  we  add  (f  )2,  =  9, 

we  have  x2  -f  6  x  +  9,  =(x  +  3)2. 

33.  By  applying  the  principle  of  the  preceding  article,  we 
can  transform  an  expression  of  the  second  degree  into  the 
difference  of  two  squares,  and  hence  factor  it. 

Ex.  l.    Factor  x2  +  6  x  +  7.  . 

We  first  complete  x*  +  6  x  to  the  square  of  a  binomial  by 
adding  (f)2,  =9.  In  order  'that  the  value  of  the  expression 
may  remain  unchanged,  we  also  subtract  9  from  it.  We  then 

have 

x*  +  6  x  +  9  _  9  +  7)  =  (3.  +  32  _  2 


Ex.  2.    Factor  x2  +  x  —  1. 

We  have          x2  +  x-l  =  x2  +  x  +       2  -       2  -  1 


Ex.  3.   Factor  3  x2  +  4  ^  -  2  f. 

Since  the  coefficient  of  x2  is  not  1,  we  first  take  out  the  fac- 
tor 3.     We  then  have 


Completing  x2  -f-  f  %y  to  the  square  of  a  binomial  by  adding 
(|  i/)2,  =-|2/2,  to  the  expression  within  the  parentheses,  and  also 
subtracting  -|  t/2  from  it,  we  have 


=  3  (x  +  |  y  +  i  V10  y)  (a  +  f  y  -  i  V10  ?)• 


250  ALGEBRA.  [Cn.  XV 

We  thus  derive  the  following  method  : 

If  the  coefficient  of  x2  is  1,  add  to,  and  subtract  from,  the  given 
expression  the  square  of  half  the  coefficient  of  x. 

Write  this  result  in  the  form  a2  —  b2  and  factor. 

If  the  coefficient  of  x2  is  not  1,  factor  out  this  coefficient,  and 
treat  the  remaining  factor  as  before. 

EXERCISES  XI. 

Factor  each  of  the  following  expressions  : 

1.  oj2  +  4aj  +  l.  2.  x2-2x-ll. 

3.  166  +  6x-ic2.  4.  9x2  +  12^-1. 

5.  4<B2-4a#  —  172/2.  6.  x2  +  %x-±. 

7.  2x2  +  6x-3.  8.  3  +  2X-11X2. 

9.  x2  —  2  mx  —  1.  10.  m-x2  —  4  mx  +  4  —  m2n. 

Evolution  of  Surds. 

34.  The  principle  established  in  Ch.  XIV.,  Art.  33,  holds 
also  for  surds.  For  any  positive  number  can  be  expressed 
as  a  power  of  a  rational  or  irrational  number,  as  in  Art.  10. 

We  therefore  have  ^/a  =  y  -tya  ; 

or,  for  present  purposes,     -\/  -\/a  =p-\/a. 
Ex.1.    ^/^/5=^/5. 
It  is  important  to  notice  that   •%/  %/a  = 
Ex.  2.    -*/  j/(S  x3)  =  j/  ^/(8  aj»)  =  j/(2  x). 
Ex.  3.         2aj  =  -2aj  x       «» 


We  thus  have  the  following  method  : 

If  possible,  take  the  required  root  of  the  radicand;  as  in  Ex.  2. 

Otherwise,  take  the  required  root  of  the  coefficient,  and  multiply 
the  index  of  the  surd  by  the  index  of  the  required  root;  as  in 
Ex.3. 

Simplify  the  result. 


33-37]  SURDS.  251 

EXERCISES  XII. 

Simplify  each  of  the  following  expressions  : 

i.  vV9-     2.  W16-       3-  W36-       4-  V(36V16)- 

5.  VV«8-  6  W«2-  7- 

s.  W(aV)-         9-  V(2ttV«2)-       10- 

11.  vVa"  12-  VV(ffaW)-  is. 

2  ..-    «-!    a 


Properties  of  Quadratic  Surds. 

35.  The  symbol  of  equality  cancelled,  ^=,  is  read  is  not  equal 
to  ;  as  2  =£  4. 

36.  ^4  quadratic  surd  cannot  be  equal  to  the  sum  of  a  rational 
number  and  another  quadratic  surd  ;  or 


wherein  y'a  and  ^/c  are  surds,  and  6  is  rational. 

For,  if  ya  =  b  +  Vc, 

then  squaring,  a  =  fr2  +  2  fry'c  +  c. 

Transposing,         2  6-y/c  =  a  —  b2  —  c. 

Dividing  by  26,       Vc  =  a  ~f  ~  °  • 

j&  0 

This  equation  asserts  that  Vc>  an  irrational  number,  is  equal 

to  -     —  —  —  ,  a  rational  number.     This  is  a  contradiction  of 

2  b 
terms,  and  therefore  the  hypothesis  ^/a  =  b  +  y'c  is  untenable. 

37.   If  a+V*  =  *+V/>  C1) 

wherein  ^/b  and  -^/y  are  surds,  and  a  and  x  are  rational,  then 
a  =  x  and  b  =y. 

For,  if  a  =£  x,  let  a  =  x  +  m. 
Then  (1)  becomes    x  -|-  m  -f  -y/&  =  a/'  +  -y/?/, 
or  m 


252  ALGEBRA.  [Cn.  XV 

But,  by  Art.  36,  this  is  impossible,  unless  m  =  0. 
When  m  =  0,  a  =x,  and  therefore  ^/b  =  ^/y. 

38.   If  V(a  +  VA)  =  V*  +  V/>  then  V(a  - 
From  y  (a  -f  yft)  = 

we  obtain  a  -f-  -^/b  =  x  -f  y  +  2 

Whence,  by  Art.  37,  a  =  #  +  y,  (1) 

and  V&  = 

Subtracting  (2)  from  (1), 


Therefore  V(a  —  V6)  =  V^  ~  V^- 

Square  Roots  of  Simple  Binomial  Surds. 

39.   Ex.  1.  Find  a  square  root  of  3  +  2^/2. 

Let  V(3  +  2  V2)  =  V*  +  Vy-  (!) 

Then,  by  Art.  38,  V(3  -  2  V2)  =  V*  ~  V2/-  (2) 

Multiplying  (1)  by  (2),  V(9  -  8)  =  a?  -  y, 
or  a;  -  y  =  1.  (3) 

Squaring  (1),  3  +  2  V2  =  a  +  0  +  2  V(*2/)  5 
whence,  by  Art-  37,  x  +  y  =  3.  (4) 

Solving  (3)  and  (4),  we  have  x  =  2,  y  =  1. 

Therefore  V(3  +  2  V2)  =  V2  +  V1  =  V2  +  L 
This  example  could  have  been  solved  by  inspection.     We 
change  3  +  2y2  into  the  form 

m  +  2-^/(mn)  +  n  =  (Vw  +  V71)2- 
We  then  have 

V(3  +  2  v2)  =  V(2  +  2  V2  +  1)  =  V(  V2  +  !)2  =  V2  +  1- 

Ex.  2.    Solve,  by  inspection,  V(21  -  3  V2^). 
We  have        V(21  -  3  V24)  =  V(21  ~  2  V54) 


37-40] 


SURDS. 


253 


In  solving  by  inspection,  first  write  the  surd  term  of  the  given 
binomial  surd  in  the  form  2^/(ww),  as  3^/24  =  2-^54. 

Then  find  by  inspection  two  numbers  whose  sum  is  equal  to  the 
rational  term  of  the  given  binomial  surd,  and  whose  product  is 
equal  to  mn. 

EXERCISES   XIII. 


Find  a  square  root  of  each  of  the  following  expressions : 

1.   7+V48-  2.  5-V24.  3.   2+V-3. 

5.  3-V5- 

8.  6  +  4V2. 

11.  11+4V?. 

I*  A-AV2- 

-  fr2).  17.    n  —  2 


6.    6+V11- 
9.    7  +  2  ViO. 
12.   30-10V5. 


4.   li+V2- 
7.   8-V28. 

10.   11-6V2. 

13.    ^ 

16.    4< 

18.    10 


Approximate  Values  of  Surd  Numbers. 

40.   An  approximate  value  of  a  surd  number  can  be  found  to 
any  degree  of  accuracy  by  the  methods  given  in  Oh.  XIV. 

Ex.  1.    Find  an  approximate  value  of  -^/2  correct  to  three 
decimal  places.     The  work  proceeds  as  follows : 


2.00'  00'  00'  00 
1_ 
100 
96 


4  00 

2  81 


1  19  00 
1  12  96 


60400 


1.4142 


24 


281 


2824 


2828 


The  work  is  simplified  by  neglecting  the  decimal  point, 
writing  it  only  in  the  result.  It  is  necessary  to  find  the  root 
to  four  decimal  places  in  order  to  determine  whether  to  take 
the  figure  found  in  the  third  place  or  the  next  greater  figure, 
according  to  the  well-known  principle  of  Arithmetic. 


254  ALGEBRA.  [Cn.  XV 

We  now  have  y'2  =  1.4142  •••. 

This  value  lies  between  1.4142,  =  yf^frf,  and  1.4143,  =  |^f . 
It  therefore  differs  from  either  of  these  fractions  by  less  than 
they  differ  from  each  other. 

"Rut  14143  _  141  42    _  1 

100TRT        linnnF  — TTFOTTO' 

Consequently  the  error  of  taking  either  1.4142  or  1.4143  as 
an  approximate  value  of  -^/2  is  less  than  JOTTO^.  By  taking 
the  root  to  more  decimal  places  a  still  more  accurate  value  can 
be  found.  It  is  therefore  possible  to  find  an  approximate 
value  such  that  the  error  will  be  less  than  any  assigned  num- 
ber, however  small. 

Ex.  2.   Find  the  value  of  -^/(l  —  x)  to  three  terms. 
The  work  proceeds  as  follows : 
l-x 


—  X 


3xl2=3 
3xl2+3xlx(— 


An  approximate  value  of  a  fractional  surd  is  obtained  most 
simply  by  rationalizing  its  denominator,  then  finding  the 
required  root  of  the  numerator  of  the  resulting  fraction,  and 
dividing  this  value  by  the  denominator. 

Q 

Ex.  3.  Find  an  approximate  value  of  — -,  correct  to  three 
decimal  places.  v 

We  have        —  =  §2^?  =  ?  (1.4142)  =  2.121. 
V2        2        2V 

EXERCISES  XIV. 

Find  an  approximate  value  of  each  of  the  following  expres- 
sions, correct  to  four  figures  : 

i.  v8-  2-  W2-5-     '       3-  V2-  4-  IV1-25- 

5.    V345-06-  6.    V10862-321-  7-    V54-0001- 

8     -?-.  9     -?_.  10        1  11     JL, 

V5  V8  2V4  V75 


IRRATIONAL  EQUATIONS.  255 


13      ^  -i-  -"-y   .  14     V-*-7 

1-V3  '    5-4VH  V2-5+V6 

Find  an  approximate  value  of  each  of  the  following  expres- 
sions, to  include  four  terms : 

18.    AV(l  +  afl.        19.    -vVfa3  -  63).        20. 


IRRATIONAL  EQUATIONS. 

41.   An  Irrational  Equation  is  an  equation  whose  members 
are  irrational  in  the  unknown  number  or  numbers  ;  as 


42.  To  solve  an  irrational  equation,  we  must  first  derive 
from  it  a  rational,  integral  equation.  This  step,  which  is 
usually  effected  by  raising  both  members  of  the  equation  to 
the  same  positive  integral  power  one  or  more  times,  is  called 
rationalizing  the  equation. 

Ex.  l.    Solve  the  equation  ^/(36  -f-  x'2)  —  x  =  2. 
Transferring  x,  V(36  +  x*)  =  2  +  x- 

Equating  squares  of  both  members, 


Transferring  and  uniting  terms, 

_4o;  =  -32. 

Dividing  by  —  4,  x  =  8. 

Check  :    V(36  +  64)  =  2  +  8,  or  ^/WO  =  10. 
Ex.  2.    Solve  the  equation  V(45  4-  ®)  +  V37  —  9- 
Transferring  -^/a;,        V(^^  +  a;)  =  9  —  ^/x. 
Equating  squares,  45  -f  x  —  81  —  18  ^/x  +  a?. 

Transferring  and  uniting  terms, 


Dividing  by  18  and  equating  squares, 

tfa*4. 

4)+V4:::::9.  or  7  +  2  =  9. 


256  ALGEBRA.  [Cii.  XV 

The  preceding  examples  illustrate  the  following  method  of 
solving  irrational  equations : 

Transform  the  given  equation  so  that  one  radical  stands  by 
itself  in  one  member  of  the  equation. 

Equate  equal  powers  of  the  two  members  when  so  transformed. 

Repeat  this  process  until  a  rational  equation  is  obtained. 

EXERCISES  XV. 

Solve  each  of  the  following  equations : 

2.   2-x  =  3.  3.   ax  =  b. 

4.    V(a;-l)  =  5.        5- 
7.   8-a:=4.  8.   9= 


25. 

26.    3  V(a?  -  3)  +  V(9  x 


5  +  3  y(a?  -7)      2  y(a?  -  7)  -  3 

' 


so. 

31.  v^  +  2tt)-V^  +  2^)  =  2V^ 

32.  V(«  +  4)  +  V(^  -  4)  =  V(4  »  -  4> 

33.  v(»  +  2)  +  V(*  -  6)  =  2  VC^  -  3)- 

34.  V(»  ~  5)  -  V(®  +  3)  =  -\/(x  -  2)  - 


CHAPTER   XVI. 

IMAGINARY   AND   COMPLEX  NUMBERS. 

1.  Attention  was  called  in  Ch.  XIV.,  Art.  10,  to  the  fact 
that  -Y/—  16  cannot  be  expressed  in  terms  of  numbers  with 
which  we  are,  as  yet,  familiar.     In  general,  since  even  powers 
of  both  positive  and  negative  numbers  are  positive,  even  roots 
of  negative  numbers  cannot  be  expressed  in  terms  of  either 
rational  or  irrational  numbers. 

It  is  therefore  necessary  either  to  exclude  from  our  consider- 
ation such  roots  as  ^/—  1,  and  in  general  Sj/—  a,  or  again  to 
enlarge  our  ideas  of  number. 

2.  We  will  now  define,  that  is,  fix  the  meaning  of,  the  num- 
bers V~~  1  an(^  A/~  a->  ky  assuming  that  they  obey  the  law 

(ya)'  =  a. 

This  relation  follows  from  the  definition  of  a  root,  as  was 
shown  in  Ch.  XIV.,  Art.  5. 

We  therefore  have  (V~  I)2  =  -  1,  and  (2-£/-  a)2n=-a. 

Whatever  meaning  and  use  these  new  numbers  have  must 
be  derived  from  these  relations. 

Imaginary  Numbers. 

3.  The  square  root  of  a  negative  number  is  called  an  Imagi- 
nary Number;  as  -^—3,  ^/—  8. 

The  study  of  these  numbers  is  simplified  by  first  considering 
the  properties  of  ^/—  1,  which  is  taken  as  the  Imaginary  Unit.* 

*  The  designation,  imaginary,  is  unfortunate,  since,  as  will  be  shown 
in  Part  II.,  Text-Book  of  Algebra,  such  numbers  are  no  more  imaginary 
(in  the  ordinary  meaning  of  the  word)  than  common  fractions  or  negative 
numbers.  Dr.  George  Bruce  Halsted,  Professor  of  Mathematics  in  the 
University  of  Texas,  has  suggested  Neomon  for  the  imaginary  unit,  and 
Weomonie  for  imaginary. 

257 


258  ALGEBRA.  [Cn.  XVI 

This  new  unit  is  commonly  designated,  by  the  letter  i,  and  its 
opposite  by  —  i. 

We  then  have  by  definition 

(y-  1)«  =  (±  /)«=-!. 

For  the  sake  of  distinction  all  numbers,  rational  and  irra- 
tional, which  have  been  used  hitherto  in  this  book  are  called 
Real  Numbers. 

4.  The  Fundamental  Operations  with  the  Imaginary  Unit.  —  It 
is  proved  in  School  Algebra,  Ch.  XX.,  that  V—  1»  or  ^  *s  used 
like  a  real  term  or  factor  in  the  fundamental  operations. 

Just  as  3  =  1  +  1  +  1,  and  -3=-l-l-l; 


—       -=----~-?  or  —=---. 


5.  We  now  have,  in  addition  to  the  double  series  of  real 
numbers,  the  double  series  of  imaginary  numbers  : 

Sit  -2  i,  -  i,  0,  i,  2  i,  3  i,  ••• . 

6.  Powers  of  /.  —  The  following  values  of  the  positive  in- 
tegral powers  of  -^/—  1,  or  i,  follow  directly  from  the  definition 
of  i  and  Art.  4 : 

V—  1  =  V—  1>  or  i  =  i, 


(V-  i)3=(V-  i)2(V- 1)  =  -  V- 1, 

(V-  I)4-  (V-  1)2( V-  1)2=  +  1,  *  =  *2  •  *2  =  4- 1, 

(V- 1)6= (V- i)4(  V- 1)2=  - 1,  ^  =  *  -?  =  - 1- 

The   preceding   results    give    the    following    properties    of 
powers  of  i: 

(i.)   All  even  powers  of  i  are  real. 
(ii.)  All  odd  powers  of  i  are  imaginary. 


3-9]  IMAGINARY  AND   COMPLEX  NUMBERS.  259 

The  sign  of  any  particular  power  of  i  is  readily  determined 
by  expressing  it  as  a  power  of  i2  if  an  even  power,  or  of  i2 
multiplied  by  i  if  an  odd  power. 

Ex.  1.  ta=(i2)11=(-l)11  =  -l. 

Ex.2.  i*=(is)18=(-l)18  =  +  l. 

Ex.  3.  i41  =  i40  x  i  =  (i2)20  •  i  =  (-  I)20  •  i  =  +  i. 

Ex.  4.  i»  =  i38  x  i  =  (i2)19  •  i  =  (-  I)19  •<»•-.  i. 

7.  Multiples  of  the  Imaginary  Unit.  —  Since 

(V-  a)2  =  -  a,  and  (V«  X  V~  I)2  =  (Va)2(V~  I)2  =  -  a, 
we  have  (V~  a)2  =  (Va  X  V  -  I)2- 

Whence  V—  a  =  Va  x  V~  *• 

Ex.i.  v~9  =  V9  x  V-1  =  3V-1  =  3*- 
Ex.2.  v-2  =  V2xV-1:=V2-  i  =  i'V2- 

In  all  reductions  involving  imaginary  terms  or  factors  it  is 
advisable  thus  to  express  them  as  multiples  of  -^/—l  or  i. 

8.  Addition    of    Imaginary    Numbers.  —  Imaginary   numbers 
are  united  by  addition  and  subtraction  just  as  real  numbers 
are  united. 

Ex.  1.    V-  9  +  V-  16  =  3  V-  1  +  4  V-  1  =  7V-  1  =  7*. 
Ex.2. 


Ex.  3.   t"  +  i15  =  i  +  (-  i}  =  0. 

9.  Multiplication  of  Imaginary  Numbers.  —  The  principle  of 
Art.  7  is  of  importance  in  the  multiplication  of  imaginary 
numbers. 

Ex.  i.  v-  9  x  V16  =  V9  x  V-  1  x  V16  =  12  V-  1  =12  i. 

Ex.2,   y-2  x  V~8=V2  x  V-l  x  V8  x  V-l 
=  V16  x(V-l)2=-4. 


260  ALGEBRA.  [Cn.  XVI 

A  point  in  Ex.  2  deserves  special  notice.     Had  we  used  the 
principle 

V«  x  y  &  = 


as  in  surds,  we  should  have  obtained 

y[(_  2)  x  (-  8)]  =  V16  =  4,  and  not  -  4. 

But  that  principle  was  proved  for  positive  roots  of  positive 
numbers,  and  therefore  cannot  be  applied  in  this  and  similar 
examples. 
Ex.  3. 

V  -  5  x  V-  10  x  V-  15  =  V5  x  V10  x  V15  x  (V-  1)3 

x  V-  1  =  —  5  i 


10.   Division    of    Imaginary   Numbers.  —  The   following    ex- 
amples illustrate  all  possible  cases. 


Ex.  3. 

V6  ye       =   yexy-i  /6      , 

_3         3x~1         ^x-l2"          3 


Ex 


V-4     V4xV-!      V4 


EXERCISES  I. 

Keduce  each  of  the  following  expressions  to  the  form  a-y/— 1, 
or  ai : 

i.  v-9-  2-  ^V-25-       3-  V~a2-      4-  aV-3^- 

5.  v-12-  6-  V-10-        7-  V-^-      a  (V 

9.  ^/(-Sa^3).      10.  V(3-27).    11.  -^/-64.     12.  -^-a12. 


0-11]          IMAGINARY   AND   COMPLEX   NUMBERS.  261 

Simplify  each  of  the  following  expressions : 
13.  i\  14.  i29.  15.  i54.  16.  £4+iM. 

17.  i-  18.  -i.  19.  4'  20- 


Add: 

21.    V-9+V-25.  22.    V-16-V-121- 

23.   v-a2-V~&2-  24.   7V-  81  +  5V-  144 

25.   5V-  8  -3V-  32.  26.   8V-  75  +V-  147. 

27.  2V-  25  -3V-  49  +  4V-  100. 

28.  2- 
29. 


Perform  the  following  indicated  operations  : 

so.  v-^4-       31-  (V-^)4-       32-  (V~a)8-       33-  V~a8- 
34.  v3xV-6-      35-  V-2xV-8-      36-  V~12xV3- 

37.    V-2xV~50-  3s-    V-axV(-9a3)- 

39.    V-^XV-^6-  40.    V~6><V12- 

41.  v-8><V-2a  42>  V-^xV-2/4- 

43.    V~2  xV~6  XV-  24.        44.    V-5xV8xV-20- 

45.  vCi-^xvO-1)-  46-  v(^~a2)xV(«-&)- 

47.    (V-5+V-3)2-  48-    (2V-3+3V-2)2. 

49.  v-3-^-v-3-      5a  V-3-^V3-       51-  V3^-V~3- 

52.    V-8-^-V-2-       53-    V-75n-v5.       54.    V12  -V-3- 

Complex  Numbers. 

11.  A  Complex  Number  is  the  algebraic  sum  of  a  real  and  an 
imaginary  number  ;  as,  3  ±  2  L 

The  general  form  of  a  complex  number  is  evidently  a  +  bi, 
wherein  a  and  b  are  real  numbers. 

When  6  =  0,  we  have  any  real  number. 

When  a  =  0,  we  have  any  imaginary  number. 


262  ALGEBRA.  [Cn.  XVI 

12.  Two  complex  numbers  are  said  to  be  equal  ivhen  the  real 
term  of  one  is  equal  to  the  real  term  of  the  other,  and  the  imagi- 
nary term  of  one  is  equal  to  the  imaginary  term  of  the  other; 
as,  2  +  3  i  =  2  +  3  i. 

That  is,  if  a  +  bi  =  c  +  <//, 

then  a  =  c,  and  bi  =  <//,  or  b  =  d. 

Observe  that  the  preceding  statement  is  a  definition  of  the 
meaning  of  the  sign  of  equality  between  two  complex  numbers. 

13.  From  the  preceding  article  it  follows  that,  if 

a  +  bi  =  0  =  0  +  0  i,  then  a  =  0,  b  =  0. 

14.  Addition    and   Subtraction    of    Complex    Numbers.  —  We 
define  algebraic  addition  of   two  or  more  complex  numbers 
as  follows  : 

Add  the  real  terms  by  themselves  and  the  imaginary  terms  by 
themselves. 


Ex.  l. 

=  _3+9V-l=-3+9i. 

15.  Multiplication  of  Complex  Numbers.  —  We  define  multi- 
plication of  complex  numbers  by  assuming  that  the  distributive 
law  holds. 

Ex.  1.      2+    3V-  1 

4-    2y-l 

8  +  12V-  1 

-  4y-i-6(y-l)2 

8+    8V-  1  +  6  =14  +  8V-  1  =  14  +  8*. 

16.  Conjugate   Complex   Numbers.  —  Two    complex   numbers 
which  differ  only  in  the  sign  of  their  imaginary  terms  are 
called  Conjugate  Complex  Numbers  ;    as, 

2  +  3V-1  and  2  -  3  V~  1,  —  4  +  6t,  -4-6t. 


12-20]      IMAGINARY   AND   COMPLEX   NUMBERS.  263 

17.  The  sum  of  two  conjugate  complex  numbers  is  real. 

The  product  of  two  conjugate  complex  numbers  is  real  and 
positive. 

Ex.  2.    (4  —  5->/—  1)  (4  4-  5^/—  1)=  42  —  (5-Y/—  I)2 

=  16  4-  25  =  41. 

18.  Division  of  Complex  Numbers.  —  We  express  the  quotient 
as  a  fraction,  and  simplify  the  result. 

Ex  1    1  +  V-  2  =  (1  +  V-  2)  (V-  3)  =  V-  3  -  y6 
2V- 3  2(V-3)2  -6 


Ex  2    —  1  _  .  _  2~V-5  _  -    2-V-5 
'  V-5)     22-(V-5)2 


19.   Any  Even  Root  of  a  Negative  Number.  —  We  have 

_  i)<  =  [(i  4V-  1)2]2 


Therefore,  -^/-4  = 

That  is,  the  fourth  root  of  —  4  is  a  complex  number. 

It  will  be  proved  in  Text-book  of  Algebra,  Part  II,  that  any 
even  root  of  a  negative  number  is  a  complex  number. 

Complex  Factors. 

20.  A  quadratic  expression  which  is  the  product  of  two 
complex  factors  can  be  resolved  into  factors  by  the  method 
used  to  resolve  a  quadratic  expression  into  irrational  factors. 

Ex.    Factor  x2-2x  +  3. 

Completing  x2  —  2  x  to  the  square  of  a  binomial  in  a?,  we  have 


264  ALGEBRA.  [CH.  XVI 

EXERCISES   II. 

Simplify  each  of  the  following  expressions  : 
1.    (2  +  4i)  +  (2i-3).  2.    (7-5z)-(3-4i). 

3.    (l+V-9)  +  (4-V-4).     4.    (6-V-l<3)-(5-V-36). 
5.    (1+V-1)(1-V-1)-       6. 
7.   (2+3V-l)(3-4V-l).     8. 

9.  (3  +  5i)(V12-30-        10.  (v8 

11.    (J-iiV3)3  +  3*3-     12.     5- 

is.    ^     ^a 


Perform  the  following  indicated  divisions  : 
14.    _  §  __         15.       -1  ___         16. 


.  . 

V-2  2-V-3  2-3V-1 


17  .          is  .         19 


1  —  i  3  —  2  i  5  —  i  V  3  a  — 

Factor  each  of  the  following  expressions  : 
21.   a58-6aj  +  26.  22.   ^  +  4*  +  68. 

23.   0^-14  a?  +  61.  24. 

25.   4»2  +  4«?/  +  3?/2.  26. 


Make  the  indicated  substitution  in  each  of  the  following 
expressions,  and  simplify  the  results  : 

27.  In  x2-  6  a;  +  14,  let#  =  3  +  y-5. 

28.  In  3^2-50;  +  7,  let  x  =  2  -  3  V~  2- 

29.  In  0^      22        i2  let  aj  =  4  +  5i      =  4  —  5t. 


CHAPTER   XVII. 

DOCTRINE   OP   EXPONENTS. 

1.  We  have  already  abbreviated  such  products  as 

act,  aaa,  aaaa,  •••,  aaa  •••  n  factors, 

by  a2,  a3,  a4,  •••,  an,  respectively,  and  called  them  the  second, 
third,  fourth,  •••,  ntl^  powers  of  a.  This  definition  of  the  sym- 
bol an  requires  the  exponent  n  to  be  a  positive  integer. 

Thus  25  means  the  product  of  5  factors,  each  equal  to  2. 
But  2°  has,  as  yet,  no  meaning,  since  2  cannot  be  taken  0  times 
as  a  factor.  For  a  similar  reason  2~5  and  2^  are,  as  yet,  mean- 
ingless. 

But,  having  introduced  into  Algebra  the  symbol  a",  it  is 
natural  to  inquire  what  it  may  mean  when  n  is  0,  negative,  or 
a  fraction. 

We  shall  find  that,  by  enlarging  our  conception  of  powers, 
quite  clear  and  definite  meanings  can  be  given  to  such  expres- 
sions as  2°,  3~2,  and  4*. 

Positive  Integral  Powers. 

2.  The  principle 

am  x  an  =  am+n, 

wherein  m  and  n  are  positive  integers,  was  illustrated  by  par- 
ticular examples  in  Ch.  III.,  Art.  24. 
In  general, 

am  x  an  =  (aaa  •••  to  m  factors)  (aaa  •••  to  n  factors) 
=  aaa  •  •  •  to  m  +  n  factors 
=  am+n. 

3.  The  other  principles  upon  which  operations  with  positive 
integral  powers   depend  have  been  proved  in  the  preceding 
chapters. 

265 


206  ALGEBRA.  [Cn.  XVII 

For  the  sake  of  emphasis,  and  for  convenience  of  reference, 
we  restate  them  here  : 

(i.)  ama"  =  am+n. 

_/W  jy  Wl 

(ii.)          —  =  am~"j  when  m  >  n  ;    -  —  =  1,  when  m  =  n  ; 
an  a" 

m  -I 

—  =  -  .  when  m  <  n. 
a"      a"~m 

(in.)  (am)"  =  am".  (iv.)  (ab)m  =  ambm. 


m        m 


Zeroth  Powers. 

4.  The  meaning  of  a  symbol  may  be  denned  by  assuming 
that  it  stands  for  the  result  of  a  definite  operation,  as  was  done 
in  letting 

an  —  a  •  a  •  a  •  •  •  •  n  factors  ; 

or  by  enlarging  the  meaning  of  some  operation  or  law  which 
was  previously  restricted  in  its  application. 

In  the  latter  way,  negative  numbers  were  introduced   by 
extending  the  meaning  of  subtraction. 

5.  We  now  enlarge  the  meaning  of  powers  by  assuming  that 
the  principle 

^=am~n 
an 

holds  also  when  m  =  n. 

We  then  have  —  =  am~m  =  a°. 

am 

But  since  —  =  1, 

am 

it  follows  that  a°  =  1. 

That  is,  the  zeroth  power  of  any  base,  except  0,  is  equal  to  1. 
E.g.,        1°  =  1,  5°=1,  99°  =  1,  (a  +  &)°  =  l,  etc. 


3-7]  DOCTRINE   OF  EXPONENTS.  267 

6.  Thus,  by  the  assumption  that  the  stated  law  holds  when 
m  =  n,  a  definite  value  of  the  zeroth  power  of  a  number  is 
obtained.     Nevertheless,  it  will  doubtless  seem  strange  to  the 
student  that  all  numbers  to  the  zeroth  power  have  one  and  the 
same  value,  namely  1.     But  it  should  be  distinctly  noted  that 

a°  is  by  definition  a  symbol  for  — ;  i.e.,  for  the  quotient  of  two 
like  powers  of  the  same  base.     Thus, 

2°=|=|=|I=1- 

Negative  Integral  Powers. 

7.  We  now  still  further  enlarge  the  meaning  of  powers  by 
assuming  that  the  principle 

an 

holds  not  only  when  m  >  n  and  m  =  n,  but  also  when  m  <  n. 
We  then  have,  for  example, 

^  =  a2-5  =  a~3. 
a5 

But,  cancelling  as  in  fractions, 

a5      a3 

Therefore,  a~s  =  —- 

a3 

In  general,  since  m  <  n,  we  may  assume  n  =  m  +  7c. 

Then  —  =  —  =  aw-<m+*>  =  a~*. 

an      am+ 

But  J£-=      1      =  1. 

am+k         am+k-m         tf 

Therefore,  a-A  =  i. 

0* 

That  is,  a  negative  power  of  a  number  is  equal  to  the  reciprocal 
of  a  positive  power  of  the  same  number,  the  exponents  being  numeri- 
cally equal. 


268  ALGEBRA.  [On.  XVII 


8.  We  also  have  —  =  —  =  ak. 

a~h      !_ 

ak 

This  relation  and  the  relation  which  defined  a  negative 
integral  power  may  be  stated  thus: 

Any  power  of  a  number  may  be  transferred  from  the  denomi- 
nator to  the  numerator,  or  from  the  numerator  to  the  denominator, 
of  a  fraction,  if  the  sign  of  its  exponent  be  reversed. 


a(—  a)4     a5 

This  reciprocal  relation  between  positive  and  negative  powers 
is  useful  in  reductions  which  involve  negative  powers. 

EXERCISES  I. 

Find  the  value  of  each  of  the  following  expressions 
1.   2-3.  2.   3-2.  3.    (I)-1.  4.    (3f)~3. 

5-    tt)-3-  6.    -^  7.    -L.  8.    (2«)-6. 

*lXt  M0 

Change  each  of  the  following  expressions  into  an  equivalent 
expression  in  which  all  the  exponents  are  positive  : 

9.   x?~4.          10.    2c~*d.          11.   3~Wn-z.      12.   5or3- 


13    2*r3  M    4^  5 

"  '  ' 


Sb~* 

In  each  of  the  following  expressions  transfer  the  factors 
from  the  denominator  to  the  numerator : 

n  9  <y3  ^  T*~3  •        'i  /r?/ 

17.    £L  18.    -££-.  19.    ££_.  20.    ^32. 

b*  5y~3  22y  ab 

3 


, 

' 


7-9]  DOCTRINE  OF  EXPONENTS.  269 

24-30.  Find  the  values  of  the  expressions  in  Exx.  17-23, 
when  a  =  3,  b  =  4,  x  =  —  2,  y  =  5. 

Fractional  (Positive  or  Negative)  Powers. 

9.  We  will  define,  i.e.,  fix  the  meaning  of,  the  power  aq,  in 
which  q  is  a  positive  integer,  by  assuming  that  it  must  obey 
the  first  law  of  exponents,  namely, 

am  -  a"  =  am+n. 

i 

In  other  words,  whatever  meaning  aq  may  have  must  be 
derived  by  an  application  of  this  law. 

By  this  law,          a?  -  a?  =  a^  =  al  =  a. 

But,  since  a*  •  a*  =  (a*)2,  by  definition  of  positive  integral 
power  of  any  base,  we  have 

(a*)2  =  a. 

That  is,  a?  is  a  number  whose  square  is  a,  or  a%  =  ^/a. 
In  general, 

1         I        I  1+M+.-.,  terms  *4 

aq  -  aq  -  aq  •••  q  factors  =  aq  9  q  =  a    q  =  a. 

111  i 

But,    since    aq  •  aq  •  aq  •••  q  factors  =  (aq)q,    by    definition    of 

positive  integral  power,  we  have  (aq)q  =  a. 

That  is,  aq  is  a  number  whose  qth  power  is  a, 

\ 
or  a<*  =  tya. 

We  are  thus  led,  by  the  definition  of  the  fractional  power, 
i 

a9,  to  the  operation  that  is  inverse  to  that  of  raising  a  number 
to  a  positive  integral  power,  i.e.,  to  the  operation  of  finding  a 
root. 

Thus,  9*  and  V9>  (-243)*  and  -fy—  243,  aq  and  ^a,  are 
only  different  ways  of  representing  the  same  numbers. 

Notice  that  the  index  of  the  root  is  the  denominator  of  the 
exponent  of  the  fractional  power,  and  the  radicand  is  the  base. 


270  ALGEBRA.  [Cn.  XVII 

10.   From  the  definition  of  a  fractional  power  we  have 

=  (V9)2  =  9,  [(-  25)*]3  =  (j/-  25)3  =  -  25. 


In  general,  (a?~)3  =  (-f/a)*  =  a. 

i 

Also,  (a9)3  =  ^/a?  =  a, 

if  only  positive  roots  be  considered. 


Therefore,  (a)9  =  (a9)?, 

for  the  positive  root. 

11.   Meaning  of  aq,  wherein  •??  is  a  positive  or  a  negative 

fraction.     We  may  always  assume  q  to  be  positive  and  p  to 
have  the  sign  of  the  fraction. 

P 
Whatever  meaning  a*  may  have   must  be   derived  by  an 

application  of  the  law 

am  •  an  =  am+n. 

By  this  law,  5*  .  5*  •  5*  =  5*+*+*  =  52. 

But,  since  5%  •  5*  -  5*  =  (5')3,  we  have  (5^)3  =  52. 

That  is,  5*  is  a  number  whose  cube  is  52  ;  or  5^  =  -%/52- 
In  general, 

tt^  •  <f  •  a*  •-  q  factors  =  a?+e+?+"'?terms==  a?'f   =  ap. 

p     p     p  P  P 

But,  since  a*  •  a*  •  a*  •  •  •  q  factors  =  (a*)7,  we  have  (a9)9  =  ap. 

That  is,  a*  is  a  number  whose  qtli  power  is  ap\ 

P 
or  ai 

Notice  that  a  fractional  power  is  a  root,  of  an  integral  power. 
The  denominator  of  the  fractional  exponent  is  the  index  of 
the  root,  and  the  numerator  is  the  exponent  of  the  power. 

•     E.g.,  23*=^234;  (-  19)*  =</(-  19)2;  2^=^2-2  =  ^/i 


10-13]  DOCTRINE   OF   EXPONENTS.  271 

12.  Since  fractional  powers  simply  afford  another  way  of 
indicating  roots,  all  the  principles  relating  to  roots  which  were 
proved  in  Chapters  XIV.  and  XV.  hold  for  such  powers. 

EXERCISES   II. 

Write  each  of  the  following  expressions  as  an  equivalent 
expression  with  radical  signs: 

1.    a*.  2.   &~*.  3.   »*.  4.   3#i 

_3     1  _5  31  -     -- 

5.   4  »""*£*  6.   2ab~5c.          7.   2~lx*y*.         8.   2  an  b  q. 

'9  «.\_J 

9.  ma-       10.  f^x  7 


xy  « 

Find  the  value  of  each  of  the  following  expressions  : 
13.   4*.  14.   169*.  15.   16~*.  16.    144~*. 

17.    27*.  18.    27~i  19.    16*.  20.   81"*. 

21.   49*  22.   512i  23.   216-*.          24.   32-*. 

Write  each  of  the  following  expressions  as  an  equivalent 
expression  with  fractional  exponents  : 

25.    -y/a.      26.  -y/a3.  27.  -^/(a~9br).        28.   -^/(2xy~5). 

29.    -Y/a2.     30.  -^/(2x~ly2).      31.  ^/(5x~y).     32.  -^/(3a~7bQ). 

13.  Having  thus  determined  definite  meanings  for  zeroth, 
negative,  and  fractional  powers,  it  remains  to  prove  that  they 
obey  all  the  principles  of  positive  integral  powers. 

Products  of  Powers. 
(I.)  aman  =  am+n, 

for  all  rational  values  of  m  and  n. 

Tjl  ,  _J5 7  ...HJ-/_7\  _.«i_7  <)  1 

Ex.  2. 


272  ALGEBRA.      ,  [Cn.  XVII 

Assume  m  to  be  positive  and  n  negative,  and  the  absolute 
value  of  m  less  than  the  absolute  value  of  n. 
Let  n  =  —  n1}  so  that  n±  is  positive.     Then 

aman  =  ama~n*  =  —  ==  —  —  =  -  1—  =  aw+<-ni)  =  am+n. 

am         ani-m         a-C«+(-ni)] 

In  a  similar  way  the  principle  can  be  proved  for  other  cases 
in  which  the  exponents  are  0  or  negative. 

That  the  principle  holds  when  the  exponents,  either  or  both, 
are  fractions,  follows  from  the  definition  of  a  fractional  power. 

EXERCISES  in. 

Simplify  each  of  the  following  expressions  : 
1.   O&E°.  2.   ar*y?.  3.   a~5a6.  4.   m~3m-5. 

5.   a8a*.  6.   cAA  7.   b~%b$.  8.   c~Vi 

9.   5a~3x3a5.       10.    -f&-2xlf&-3.     11.    a86-*  X 

12     12«-3x.    a2  7c-8  .  35  a-4 

n-2        9n3  3  a3   '     6c2   ' 

15.    (at4-«"*)(a   —«"*)•  16.    (a*  -f  a 

17.     a^-c^^  +  ^a^-f^.          18. 


20.  (aj»  -  x-3  -  2  x-6  +  5)  (10  x-7  +  x~l- 

21.  (.^  —  a^  +  x^y  —  y%)  (x  -f  a^  +  y). 

22.  (a*  +  a~*  —  a*  -  a~^)  (a*  +  a~*  +  1). 

23.  (»*  +  2x-5  +  3»*  +  2^  +  1)  (^  -  2»*  +  1). 

Quotients  of  Powers. 

(ii.)  J  =«--  , 

for  all  rational  values  of  m  and  n. 

Ex.  1.  -^L  = 

or3 


13]  DOCTRINE  OF   EXPONENTS.  273 

Ex.2.  £^  =  o«&M  =  a-*&*=^ 

c&T*  a* 

We  have  —  =  ama~n  =  am+(~n)  =  aw~n. 

an 

EXERCISES   IV. 

Simplify  each  of  the  following  expressions : 


1.  JL. 

2.   Jl.. 

K-2 

3-!k 

4'T 

i.g. 

•£ 

7'^' 

8.    —  • 

x~n 

9.   ££H!. 

orn 

10-S 

11.    (!£&-*)-!- (3  fc2).  12.   1- 

13.    (3J  an&-4)  -T-  (J  aw&-3).  14.    (a^  —  &*)  -:-  (a^  +  6*). 

16.  (3  a-10  +  a6  -  4  a~6)  -s-  (2  a~2  +  a2  +  3  a~6). 

17.  (2  x-3  -  3  or2  -  2  x-1  +  2  -  a)  -  (ar1  +  1). 

19.  (2  a7  -  3  a3  -  23  a-1  +  15  a~5  +  9  or*)  •*-  (a4  +  2  -  3  a~4). 

20.  (6a^  +  9a;~*  -  2X-1  -  13)  -5-  (3 a;*  +  2x~^  -  5). 

22.    (a?  —  «262  _  cfiW  +  6*)  -s-  (a*  —  a&^  4-  ft *b  —  &T). 

Powers  of  Po\nrers. 

(III.)  (am)'7  =  a""7, 

for  all  rational  values  of  m  and  n. 

Ex    1  /Vy.2\-3 y2(— 3)  y-6 _£_. 

£C6 

Ex.  2.  (1024*)^  =  1024-*  =  /10;,1non3  =  |. 


274  ALGEBRA.  [Cn.  XVII 

(i.)    m  and  n  both  negative  integers. 

Let  m  =  —  ?%  and  n  =  —  nlt  so  that  mj  and  ^  are  positive. 

We  have 

(am)*  =  (a-TO1)-ni  =  f-L     "1=  (ami)M1  =  awini  =  a(-TOl)(-ni)  =  amB. 


In  a  similar  manner  the  principle  can  be  proved  for  other 
cases  in  which  the  exponents  are  0  or  negative  integers. 

(ii.)    m  a  fraction,  and  n  a  positive  or  a  negative  integer,  or  0. 
Let  m  =  :?,  wherein  q  is  a  positive  integer  and  p  is  a  posi- 

tive or  a  negative  integer. 
We  then  have 

p  1  1  pn  p^ 

(am)n  =  (a*)"  =  [(«*)p]n  =  (a*)1"1  =  a  ?  =  a9  =  amn. 

In  a  similar  manner  the  principle  can  be  proved  when  m  is 
an  integer  and  n  is  a  fraction. 

(iii.)   m  and  ?i  both  fractions.     Let  m  =  :?,  and  M  =  1- 

q          s 


p 


If  (a9)*  be  raised  to  the  gsth,  =  sqth  power,  we  have 


?  r 
Consequently  (a?)s  is  the  qs  root  of  apr;  or,  by  definition 

of  a  fractional  power, 

^  r  pr  p    r 

(o?)'  =  «••==  a*'*. 

EXERCISES   V. 

Simplify  each  of  the  following  expressions  : 
1.  (x2)-2.  2.  (a8)*  3.  [(-a)*]8.         4.  (a?-8)4. 

5.   (offy5.  6.   (a-3)*.'  7.   O3)^.  8.  (x-2)-5. 

p    n> 

9.  (af^fi          10.  (aM)-2.  11.  (a-M)-3.  12.  (a~«)"». 

13.         a-24.         14.         «~f-         15- 


13]  DOCTRINES   OF   EXPONENTS.  275 

Powers  of  Products. 

(IV.)     (ab)m  =  ambm,  for  all  rational  values  of  m. 

Ex.1.    (2  a)-  =  2-**-  =  ^. 

Ex.  2.    (3  of  Vr4  =  3-Vsr8  =  £j—  8- 

(i.)  m   a  negative   integer.     Let  m  =  —  m1?  so   that   mj  is 
positive. 

Then  (ab}m  =  (ab}~mi  =  -^—  =  —  i—  =  a~mib-'ni  =  ambm. 
(a6)OTi      arai&rai 

(ii.)  m  a  fraction.     Let  m  =  —  ,  where  j9  is  a  positive  or  nega- 

tive integer,  and  q  is  a  positive  integer. 

P 
If  («6)?  be  raised  to  the  gth  power,  we  have 

[(a&)*]?  =(a6)p,  since  q  is  an  integer, 
=  apbp,  by  (i.). 

But  (V>~)«  =  (a«~)«  (6?")9  =  a^6p. 

Therefore  [(a&)*]«  =(a*6*)ff  ;  whence  (aft)*  =  a«b«. 

Powers  of  Quotients. 

(V.)          [  -  )    =  —  ,  for  all  rational  values  of  m. 
o          bm 


i 


We  have  -      =  (ab~l)m  =  qmb-m  = 

EXERCISES   VI. 

Simplify  each  of  the  following  expressions  : 
1.    (aV1)-2.  2.    (la)-'.  3.    (8a-6)i 

4.     a-^-8-4.  5.     2a*aj*  6.     ajV*' 


276  ALGEBRA.  [Cn.  XVII 


7.    ^-T    •  a       r "      •       9. 


W     J  \f    ) 

10.    f       ""     J   '•  11.    f-^LV5-  12. 


13.       -^-1     -  14.       ^-;     •  15. 


EXERCISES   VII. 
MISCELLANEOUS    EXAMPLES. 

Simplify  each  of  the  following  expressions : 

a  —  b  _  a?  —  b\  2      a  —  x         a  4-  x 

^1/ri  _L  /it/r^        n        w  ^  _L  1  1 

a^  ^    ~r  a/  tC                    ^                       A          a;"  ~r  •»-         .  J- 

*. r" 


1          .,  ^1-5         -I 

a?  4-  x*  4  1      x 


1^-+ 


Find  the  square  root  of  each  of  the  following  expressions 
6.    x?  -f  a;~^  +  2.  7.    a-4x  +  2  a~  V^  +  aa;-4. 

8.  4  it--4  -  12  or3  +  13  or2  -  6  or1  4- 1. 

9.  9x2  +  10ar2-4ar4  +  ar6-12. 

10.  a2  -  f  a^  -  f  a*  +  f  1  a  4  1. 

11.  f  aj8  -  5  a; V  +  -W-  ^  -  I  *¥  +  A  ^2- 
Find  the  cube  root  of  eacn  of  the  following  expressions : 
12.   ar6-6or54l2ar4-8ar3.     13.    8^-36^-27^4  54 
14.    oj*  -  3  a?*  4-  3  a?*  +  2  x  +  3  a?*  -  3  »*  -  6  o?^  4 


CHAPTER   XVIII. 

QUADRATIC   EQUATIONS. 

1.   A  Quadratic  Equation  is  an  equation  of  the  second  degree 
in  the  unknown  number  or  numbers. 


E.g.,         x2=25,  x2 

A  Complete  Quadratic  Equation,  in  one  unknown  number,  is 
one  which  contains  a  term  (or  terms)  in  x2,  a  term  (or  terms) 
in  x,  and  a  term  (or  terms)  free  from  x,  as  x2—  2ax-\-b=cx—  d. 

A  Pure  Quadratic  Equation  is  an  incomplete  quadratic  equa- 
tion which  has  no  term  in  x,  as  x2  —  9  =  0. 

Pure  Quadratic  Equations. 

2.  Ex.  l.    Solve  the  equation  6  or2  —  7  =  3  or2  +  5. 
Transferring  3  x2  to  the  first  member,  and  7  to  the  second 

member,  6^-3^  =  5+7, 

or  3  a2  =  12. 

Dividing  by  3,  x2  =  4. 

The  value  of  x  is  a  number  whose  square  is  4.     But 
22  =  4,  and  (-2)2  =  4. 

Therefore  x  =  ±2. 

3.  This  example  illustrates  the  following  principle,  which  is 
proved  in  School  Algebra,  Ch.  XXI.  : 

The  positive  square  root  of  the  first  member  of  an  equation  may 
be  equated  in  turn  to  the  positive  and  to  the  negative  square  root 
of  the  second  member. 

277 


278  ALGEBRA.  [Cn.  XVIII 

Ex.  2.    Solve  the  equation  (a;  —  2)  (x  +  2)  =  —  6. 
Simplifying,  x2  —  4  =  —  6. 

Transferring  —  4,  cc2  =  —  2. 

Equating  square  roots,  a;  =  ±  ->/—  2. 

These  results  are  imaginary.     Yet  they  satisfy  the  given 
equation,  since 


In  such  a  case  the  equation  is  said  to  have  imaginary  roots. 
The  meaning  of  an  imaginary  result,  when  it  arises  in  connec- 
tion with  a  problem,  will  be  explained  in  Art.  16. 

4.  The  methods  used  in  Ch.  VIII.  for  solving  fractional 
equations  which  lead  to  linear  equations  apply  also  to  frac- 
tional equations  which  lead  to  quadratic  equations. 

Ex.  3.    Solve  the  equation  ^-±-^  +  ^-^  =  0. 

b  -f  x     x  —  b 

Clearing  of  fractions, 

(a  +  x)(x  -  b)  +  (a;  -  a)  (b  +  x)  =  0, 
or,  x2  4-  ax  —  bx  —  ab  -f  x2  —  ax  +  bx  —  ab  =  0. 

Transferring  and  uniting  terms, 


Dividing  by  2  and  equating  square  roots, 

x  =  ±  y  (a&). 
This  equation  therefore  has  irrational  roots. 

EXERCISES   I. 

Solve  each  of  the  following  equations  : 
1.   a2  =  729.  2.   x2-  25  =  144.       3.   5  a;2-  27-=  2  or2. 

3_x  ?_^_JL  g    a?2  —  1_0 

"      ~'  '         ~ 


10.   7aj2-8  =  9aj8-10.  11.   5  +  16  a2  =  11  x2  +  15. 


3-4]  QUADRATIC   EQUATIONS.  279 


12. 

13.   5(3arJ  +  l)  +  81  =  7(5a;2-16)  +  18. 

14    A     JL  =  JL  15    2-x2     7ar*  +  9          37 

'   2x2     3x2     12  5  6  15* 

__    „     15  — x     A  .  #  +  10 

16.     f —  =  D  H •          17. 

ar  ar 

18.    (7 +  2 a;) (7 -2 a;)  =13.  19.   (a?  +  i)(a;  -  J)  =  . 

20.    0-8)0+5)=3(3-#).  21.    0  +  2)0+3)  =5 

22.    (x  +  3)2  =  49.  23.    (3  x  +  4)2  —  49  = 

24.   64  x2  —  80  x  +  25  =  9.  25. 


26.  '         =          '      .  27. 


/y  j_  P»  ^^99^^ 

»*/   — J"~  *J  t_/  *v  —  —  «#  «C  O 

28     ^±5_  =  1^±1.  29    3a?-4  =  7a;4-24. 

ic  +  13     3a;  +  18  '   4^-1      8^  —  19 

.T  +  3        10        1  6a;     14  +  a^_o 

3a  —s — ^Ti  =  2*  31'  T"2^T7  = 

32.  (2  x  -  3)  (3  a;  -  4)  — (a;  -  13)0  ~  4)=  40- 

33.  (5  x  -  7)  (3  x  +  8)  -  0  -  10)  (9  -  a?)  =  1634. 

34.  a?-l   [  a?  +  l 


35. 
36. 


07  +  1        #  —  1          X2  —  1 

5  3  •  132 


x  -  5     2  x  +  3     77  (a;  -  6) 
64  11  6  81 


x  +  7      a; -8     a;  +  2     a;  +  12 

37.  (5  -  a?)  (3  -  »)  (1  4-aj)4-(54-aj)(3  +  aj)(l  -  a?)  =  16. 

38.  ax2  =  64.                                      39.    (a  —  bx)2  =  c2. 
40    ax2  -4-  b2  =  bx2  +  a2  41     (x  4-  CL](X a] • 


42.   m2x2  —  4  mx  +  4  =  9.  43 ,          _„    , 

a  b 

44.    -JL-  +  -?L-  =  1.  45.         "2  *2 

a  +  x     6  +  x 


ax  —  b  _  bx  +  a  47    a;  +  l  __  a  +  bx  +  car^ 

a  —  bx      b  +  aa;  a;  —  1      a  —  to  +  car1 


280  ALGEBRA.  [Cn.  XVIII 

Solution  by  Factoring. 

5.  The  principle  on  which  the  solution  of  an  equation-  by 
factoring  depends  was  proved  in  Ch.  VI,  Art.  43.  The  methods 
given  in  Ch.  VI,  Arts.  9-13  ;  Ch.  XV,  Art.  33  ;  and  Ch.  XVI, 
Art.  20,  enable  us  to  factor  any  quadratic  expression.  The 
roots  of  the  given  quadratic  equation  are  the  roots  of  the  equa- 
tions obtained  by  equating  to  0  each  of  its  factors. 


Ex.  1.   Solve  the  equation 

Dividing  by  3,  x2  +  f  x  -  f  =  0. 

Adding  and  subtracting  (.jl^)2,  =  ff>  we  have 


or,  (*  +  f)2-M  =  0. 

Factoring,  (x  +  f  +  f)  (x  +  f  -  J)  =  0, 

or,  (x  +  2)(x-®=0, 

Equating  each  factor  to  0, 

x  +  2  =  0,  whence  x  =  —  2  ; 
x  —  i  =  0,  whence  x  =  -|. 

Ex.  2.   Solve  the  equation  2a2  +  2#-l  =  0. 

Dividing  by  2,  x*  +  x  —  %  =  0. 

Adding  and  subtracting  (i)2,  =  J, 


or  *  + 

Factoring,  (x  +  i  +  i  V3)  0»  +  i  -  iV3)  =  °- 
Equating  factors  to  0, 

a  +  i  +  lV3  =  °>^+-i-  i  V3  =  °- 
Whence  x  =  -  |  -  £  ^/3,  and  -  £  + 
Such  roots  are  usually  written  —  J  ± 


5]  QUADRATIC   EQUATIONS.  281 

Ex.  3.   Factor  x2  -  2  x  +  19  =  0. 
Adding  and  subtracting  (—  I)2,  =  1, 

x*-2x  +  l  -1  +  19  =  0, 

or,  since  -  1  +  19  =  18  =  -  (-  18)  =  -  (V~  18)2, 

=  -(3V-2)2, 
0-1)2-(3V-2)2  =  0. 

Factoring,    (a;  -  1  +  3  V-  2)  (a;  -  1  -  3  V-  2)  =  0. 
Equating  factors  to  0, 

x  -  1  +  3  V-  2  =  0,  a>  -  1  -  3  V-  2  =  0. 
Whence,  x  =  1  ±  3  v'  -  2. 

EXERCISES  II. 

Solve  each  of  the  following  equations  : 

1.  x2-  6#-f-  5  =  0.  2.   0^-7^  +  10  =  0. 

3.  ie2-4a;-21=0.  4.   ^  =  11^  +  12. 

5.  3z2  +  4z  +  l  =  0.  6.   9  a*  -12  a?  +  4  =  0. 

7.  6x2  +  13z-8  =  0.  8.    11  a2  -7^-18  =  0. 

9.  7a2-20a  +  8  =  0.  10.   7-12a?2  =  17aj. 

11.  20^-79^  +  77  =  0.  12.   8x2  +  13a;-82  =  0. 

13.  a,-2-  2  cc-  1  =  0.  14.   or8  —  6  a?  -71  =  0. 

15.  x2-2x  +  2  =  0.  16.   a2-  4^  +  13  =  0. 

17.  (a?  +  8)(aj  +  3)=aj-6.  18.    (x  +  7)  (x  -  7)  =  2(a?  +  50). 

19.  (2a  +  l)0  +  2)  =  3a2-4  20.    (aj-l)(2a?  +  3)=4aj2-22. 

21.  .T2-3  =  i(x-3).  22.   x(^  +  5)  =  5(40-a^)  +  27. 

23          ^                14  24      x  +  7  =3a;-5 

x  +  120     3  x  -  10  2  2  +  3       x  +  3  ' 


25  5     =a?  +  ll.      26    5     4a?  +  7=      3, 

4         a;  —  6          6  '   x       x  +  1  "        2 


27     __      __       __          28 

' 


282 

ALGEBRA. 

[Cn. 

XVIII 

29. 

Qx-Sx2 

x 

x+ 

3 

30.    — 

X- 

-1 

4 

1 

1 

21-7x 

21 

x 

x-2 

x 

— 

3 

31. 

1  x2  -  106 

1 

32   X~ 

-  2 

+ 

x  4-  2 

2 

(x 

+  3) 

x  +  3 

8^-72 

8 

x  H 

-2 

a?  -2 

x-3 

33. 

x  +  24 

x-7 

1 

34      4 

x-  +  67 

x 

2 

5  ar2  -  5      3.  +  i      2  a?  - 
35.    ar}  +  llaz  +  28a2  =  0.  36.    x2  -  14  mx  +  33  m2  =  0. 

37.   x2  —  2  ax-{-  a2—  b2  =  0.  38.   a?2— 3  a#-f  2 a2— ab  —  62=0. 

39.  «2  -  (2  m  -  1)  a?  +  m2  -  m  -  6  =  0. 

40.  x2  -  (3  a  +  2  6)  a;  +  6  06. 

41.   ax2  +  (a  +  2)  a  +  2  =  0.          42.   bx2  -  2  (b  +  c)  x  +  4  c  =  0. 

43.  (a  + 

44.  (a2-| 

45.  a2  — 

46.  (m  —  w)  x2  —  (m  -f  w)  #  +  2  n  =  0. 


47.    _        +  __  =  2.  48. 


. 

—  ox  —  a  2  a  —  b 


a  ( 


a?      a6  (6  —  1)      6  #  +  4  >i      x  —  4  ?i 

Solution  by  Completing  the  Square. 

6.  The  following  examples  illustrate  the  solution  of  a  quad- 
ratic equation  by  the  method  called  Completing  the  Square. 

Ex.  1.    Solve  the  equation  a2  —  5  x  +  6  =  0. 

Transferring  6,  x2  —  5  x  =  —  6. 

To  complete  the  square  in  the  first  member,  we  add  (—  f)2, 
=  -^-,  to  this  member,  and  therefore  also  to  the  second.  We 
then  have 


Equating  square  roots,  x  —  -f  =  ±  £,  by  Art.  2. 


5-6]  QUADRATIC   EQUATIONS.  283 

Whence,  x  =  f  ±  £. 

Therefore  the  required  roots  are  3  and  2. 
Ex.  2.    Solve  the  equation 


Transferring  1,  7  x2  +  5  x  =  —  1. 

Dividing  by  7,  #2  +  f  x  =  —  f  . 

Adding  (^Y  =  AV,  ^  +  *  *  +  T¥*  =  fWr  -  *  =  f 
Equating  square  roots,  x  -f  T\  =  ±  TW~  3- 
Whence,  x  =  —  T\  ± 

Therefore  the  required  roots  are 

-3  and  -       - 


Ex.  3.   Solve  the  equation 


Transferring  a2,  (a2  -  62)  x2  - 

Dividingby^,  rf- 


*°  ^  members> 


a4  a2          a^ 


Equating  square  roots,  x 2a       =  ±  -^ — 

Whence,  *  =  ^^- 


Therefore  the  required  roots  are      a      and 


a  —  b          a  +  b 

The  preceding  examples  illustrate  the  following  method  of 
procedure : 

Bring  the  terms  in  x  and  x2  to  the  first  member,  and  the  terms 
free  from  x  to  the  second  member,  uniting  like  terms. 


284  ALGEBRA.  [Cn.  XVIII 

If  the  resulting  coefficient  of  x2  be  not  +  1,  divide  both  members 
by  this  coefficient. 

Complete  the  square  by  adding  to  both  members  the  square  of 
half  the  coefficient  of  x. 

Equate  the  2^ositive  square  root  of  the  first  member  to  the  posi- 
tive and  negative  square  roots  of  the  second  member. 

Solve  the  resulting  equations. 

EXERCISES  III. 
Solve  each  of  the  following  equations : 

0.  2.   a*-5a;  =  -4. 

3.   a2 +  2^  +  1  =  0.  4.    2  or2- 

5.   3a2-53a  +  34  =  0.  6. 

7.   x2-4z  +  7  =  0.  a 

9.   x2-2x  +  6  =  0.  10.   x2-l-^-x(x- 

11.    (3x-2)(x-l)  =  U.        12.   (2a-l)(a-2) 

13.   #  +  -  =  54-.  14.   x--=U. 

x  x 

15.   £_!=:!?.  16.    —  =  a;-4. 

x  x 

17.    -i-  +  J-=z-i.  18.    a+i=7+i. 

2x     3x  6  *  7 

19.    — ^=a;  +  2.  20.    2 


21.    a?  +  3=      a;-4  22    a?+l  =  3a?  + 

x  +  9         cc  — 1  '   aj-ho     7^  — 

23        1Q      |        27      =5.  24.    ^+j_2^~ 

1  —  x     1  —  2ic  re  — 5       x  + 

25.  (2x-3)2  =  8x.  26.    (2 

27.  (5z-3)2-7=40a-47.       28.    (x 

29.  (aj~7)(aj-4)  +  (2aj-3)(aj-5) 

30.  10(2  a?  +  3)  (a?  -  3)  +  (7  «  +  3)2  =  20  (x  +  3)  (a;  -  1). 

31.  (a?  -  1)  (a;  -  3)  -f-  (x  -  3)  (or  -  5)  =  32. 

32.  (x  -  1)  (a?  -  2)  +  (x  -  3)  (a?  -  4)  =  (x  -  I)2  -  2. 


6-7]  QUADRATIC   EQUATIONS.  285 

638 


33. 


34. 


ic  —  5     x  —  4     x  —  3 
12  7  15 


-1     6  —  x         x-2 
35.    -^  +  -6+^  =  5. 


36    2x-7_       7  11 

9  ~     1       "K          "A       o 

«7  3  ic          ,  6 


38. 


-f  3 
2 


39.  |  !+*  l-3a? 


40.   aj_     =    -.  41.  |      -    =  . 

6     a     a?  ?i  —  a;     7i  -f  x     n2  —  x2 

42.    a.=  • 


" 


(a  —  6)2x     a  —  6  «2a;  —  2  H     2  —  no;     ?i 

1  a  a-1 


2  a 


—     —  a?  «a;  —  a 


a  -f  6  +  x  CKC  + 1      ^2ic2  —  <*  —  a2#H-ao? 

48. 


a  —  x        2     a  —  x 


General  Solution 

7.   The  most  general  form  of  the  quadratic  equation  in  one 
unknown  number  is  evidently 

as?  +  bx  +  c  =  0. 

The  coefficient  a  is  assumed  to  be  positive  and  not  0,  but  b 
and  c  may  either  or  both  be  positive  or  negative,  or  0. 


286  ALGEBRA.  [On.  XVIII 

Dividing  by  a,  x2  +  -  x  +  -  =  0. 

a        a 

Transferring  -,  y?  +  -  x  =  -  -. 

a  a  a 

Adding  (AT-  =  &' 


4  a2 
Equating  square  roots,  a  +  —  =  ±  V(&2~4ac). 

Whence,  x 


2a  2a 

and  a  =  - 


2ci  2a 

8.  The  roots  of  any  quadratic  equation  can  be  obtained  by 
substituting  in  the  general  solution  the  particular  values  of  the 
coefficients  a,  b,  and  c. 

Ex.   Solve  the  equation  3ot?  +  7  x  —  10  =  0. 

We  have  a  =  3,  6  =  7,  c  =  -  10. 

Substituting  these  values  in  the  general  solution,  we  obtain 

*  =  -£  +  iV[49-4x3(-  10)]  =  1, 
and          •      a  =  -i 


EXERCISES  IV. 

Solve  each  of  the  following  equations  : 
1.   2a2  =  3o;  +  2.  2.   5a^-6a;  +  l  =  0. 

3.   9o;(a?  +  l)  =  28.  4.   x2-b2  =  2ax-a? 

5.   oj2  +  6oa;  +  l=0.  6.   ar2  +  l  =  2i#. 

7.    (a-5)2+(>-10)2  =  37.       a    2tt(3n-4a;)  =  tt2 
9.   n2(»2  +  l)  =  a2  +  2n2a?.         10.   x2  +  (a?  +  a)2  =  a2. 


7-10]  QUADRATIC   EQUATIONS.  287 

Relation  between  Roots  and  Coefficients. 
9.   If  the  roots  of  the  quadratic  equation 

ax2  +  bx  +  c  =  0,  or  x2  -}--a;  +  -  =  0 
a        a 

be  designated  by  r±  and  r.2,  we  have 


r  =       b       V(62-4ac) 

2a  2a 

The  sum  of  the  roots  is 

n+r,=-£-  (i) 

The  product  of  the  roots  is 
b 


-4ac)-]      r       6 
T~     J     L~^~ 


The  relations  (1)  and  (2)  may  be  expressed  thus  : 

(i.)  If  the  coefficient  of  the  second  power  of  the  unknown  num- 
ber be  1,  the  sum  of  the  roots  is  equal  to  the  coefficient  of  the  first 
power  of  the  unknown  number,  with  sign  reversed. 

(ii.)  If  the  coefficient  of  the  second  power  of  the  unknown  num- 
ber be  1,  the  product  of  the  roots  is  equal  to  the  term  free  from  the 
unknown  number. 

E.g.,  the  roots  of  the  equation  x2  —  5#  +  6  =  0  are  2  and  3  ; 
their  sum  is  5  (the  coefficient  of  x  with  sign  reversed),  and 
their  product  is  6  (the  term  free  from  x). 

The  roots  of  the  equation  6  x2  —  x  —  2  =  0,  or  x2  —  J  x  —  ^  =  0, 
are  J  and  —  ^  ;  their  sum  is  -J,  and  the  product  is  —  ^-. 

10.  Formation  of  an  Equation  from  its  Roots.  —  The  relations 
of  the  last  article  enable  us  to  form  an  equation  if  its  roots  be 
given.  We  may  always  assume  that  the  coefficient  of  the 
second  power  of  the  unknown  number  is  1. 


288  ALGEBRA.  [Cn.  XVIII 

Ex.  1.   Form  the  equation  whose  roots  are  —  1,  2. 
We  have  r±  -f  r2  =  —  1  4-  2  =  1,  the  coefficient  of  x,  with  sign 
reversed  ;  and  rfa  =  —1  x  2  =  —  2,  the  term  free  from  x. 
Therefore  the  required  equation  is  x2  —  x  —  2  =  0. 

Ex.  2.   Form  the  equation  whose  roots  are  1  -f  2^/3,  1  —  2^/3. 
We  have  i\  +  r2  =  (1  +  2  V3)  +  (1  -  2  V3)  -  2  ; 
and  rfa  =  (1  4-  2  V3)  (1  -  2  y3)  =  1  -  12  =  -  11. 

Therefore  the  required  equation  is  cc2  —  2  x  —  11  =  0. 

11.   It  follows  from  Art.  9,  that  the  quadratic  equation  may 
be  written  in  the  form 

^-Oi  +  r^aj  +  r^^O, 
or  (x  —  rx)  (x  —  r2)  =  0. 

Ex.   Form  the  equation  whose  roots  are  —  1,  2. 
We  have    (x  -f  1)  (a;  —  2)  =  0,  or  x2  -  x  -  2  =  0. 

When  the  roots  are  irrational  or  imaginary,  the  method  of 
the  preceding  article  is  to  be  preferred. 

EXERCISES   V. 

Form  the  equations  whose  roots  are  : 

1.   8,  2.              2.    -  5,  -  3.        3.    10,  10.  4.  7,  -  3. 

5.   4,  -  10        6.   2£,  If.             7.    -  j,  -  1  J.  8.  -  J,  8. 

9.    2,  0.            10.    a,  b.               11.    -  a,  -  1.  12.  a2,  -  4  a2. 

is.  v2>  -V2-  14-  W~3>  -4V-3- 

15.   1  -}-  V7,  1  -  V7.  16.   i  -  i  V11*  *  +  t  V11- 

17.   3--5   3+-5.  18.      - 


Nature  of  the  Roots. 

12.  In  many  applications  it  is  important  to  know,  without 
having  to  solve  an  equation,  the  nature  of  its  roots,  i.e.,  whether 
they  are  both  reed  and  unequal,  whether  they  are  both  real  and 
equal,  whether  they  are  imaginary. 


10-12]  QUADRATIC   EQUATIONS.  289 

In  the  general  solution 

b       y(62-4ac)  b       y(&2-4ac) 

2a  +          2a  2a          ~~2^~ 

of  the  equation  aa?2  +  b%  +  c  =  0, 

a,  6,  and  c  are  limited  to  real,  rational  values. 

(i.)   The  tivo  roots  are  real  and  unequal  when   b2  —  4  ac   is 
positive,  i.e.,  when  b2  —  4  ac  >  0. 

E.g.,  in  x2  +  4#  -  12  =  0, 

a  =  1,  6  =  4,  c  =  —  12  ;  and  since  62  —  4  ac,  =  16  +  48, 
is  positive,  the  roots  of  this  equation  are  real  and  unequal. 

(ii.)   The  two  roots  are  real  and  equal  when  b2  —  4  ac  is  equal 
to  0  ;  i.e.,  when  b2  =  4  ac. 

E.g.,m  orj-4x  +  4  =  0, 

a  =  1,  b  =  —  4,  c  =  4  ;  and  since  62  =  4  ac, 
the  roots  of  this  equation  are  real  and  equal. 

(iii.)   The  two   roots  are   conjugate    complex    numbers    when 
b2  —  4  ac  is  negative  ;  i.e.,  when  b2  —  4  ac  <  0. 

E.g.,  in  ^-2^  +  3  =  0, 

a  =  1,  b  =  -  2,  c  =  3;  and  since  b2  -  4  ac,   =  4  -  12,  =  -  8, 
is  negative,  the  roots  of  this  equation  are  complex  numbers. 

EXERCISES  VI. 

Without   solving  the   following   equations,   determine    the 
nature  of  the  roots  of  each  one  : 


=0.     2.  a^+12  z=  -40.     3. 
4.   x2-x  =  12.  5.   x2-  Sx  +  25  =  0. 

6.   aj2-8»  =  16.  7. 

8.   8^-2^-25  =  0.  9. 

10.    10  a2-  21  a;  -10  =  0. 


290  ALGEBRA.  [Cii.  XVIII 

For  what  values  of  m  are  the  roots  of  each  of  the  following 
equations  equal  ?  For  what  values  of  m  are  the  roots  irra- 
tional? And  for  what  values  of  m  are  the  roots  complex 
numbers  ? 


11.   wa;2-h4a,'  +  l=:0.  12.    2ar>  -f  mz  +  1  =  0. 

13.   3#2  +  6o;  +  m  =  0.  14.   mo?  +  wa  +  1  =  0. 

IRRATIONAL  EQUATIONS. 

13.   An  irrational  equation  may  lead  to  a  quadratic  equation 
when  rationalized. 


Ex.  l.    Solve  the  equation  x  +  ^/(25  —  x2)  =  7. 
Transferring  x,      V(25  —  ^2)  =  7  ~  ^  0-) 

Squaring,  25  -  or2  =  49  -  14  oj  +  x2.  (2) 

The  roots  of  this  equation  are  3,  4. 

Both  roots  of  (2)  satisfy  the  given  equation,  since 

3  +  V(25  -  9)  =  7,  and  4  -f-  V(25  ~  16)  =  7. 

Ex.  2.   Solve  the  equation  a?  —  V(25  —  a2)  =  1. 
Transferring  #,     —  -^(2o  —  x2)  =  1  —  #.  (1) 

Squaring,  25  -x2  =  1  -  2x  +  x2.  (2) 

The  roots  of  this  equation  are  4  and  —  3. 

The  number  4  is  a  root  of  the  given  equation,  since 

4-V(25-16)  =  l; 

but  the  number  —  3  is  not  a  root  of  the  given  equation,  since 
-  3  -  V(25  -  9)  =  -  7,  not  1. 

Therefore  the  root  —  3  was  introduced  by  squaring.  Now 
observe  that  the  same  rational  equation  (2)  would  have  been 
obtained,  if  the  given  equation  had  been 


that  is,  if  the  surd  term  had  been  of  opposite  sign.     The  root 
—  3  satisfies  equation  (3),  since 

-  3  +  V(25  -9)  =  -3  +  4  =  1. 


12-13]  IRRATIONAL  EQUATIONS.  291 

Therefore  equation  (2)  is  equivalent  to  equations  (1)  and  (3) 
jointly. 

It  frequently  happens  that  no  root  can  be  found  to  satisfy 
an  equation  obtained  by  giving  to  the  square  root  either  its 
positive  or  its  negative  value. 

In  Ex.  1,  the  equation  thus  derived  is 


and  is  not  satisfied  by  either  of  the  roots  obtained.     The  equa- 
tion is  then  said  to  be  impossible. 

Ex.  3.     Solve  the  equation 

V(2*  +  3)  -V(7  -*)  =  !. 

If  both  positive  and  negative  square  roots  be  admitted,  the 
given  equation  is  equivalent  to  the  four  equations  : 

-s)  =  l  CO,  ^(2x+3)-^/(7-x)  =  l  (2), 
-x)  =  l  (3),  -V(2a+3)-  V(7-®)  =  1  (4). 

The  same  rational  integral  equation  will  evidently  be  de- 
rived by  rationalizing  any  one  of  these  equations. 

In  (1)  transferring  -^/(7  —  x), 


Squaring,       2  x  +  3  =  1  - 

or  3o;  —  5  =  — 

Again  squaring,  9  x2  —  30  x  +  25  =  28  —  4  x, 
or  9aj8-26a?-3  =  0. 

The  roots  of  this  equation  are  3  and  —  i-.  By  substitution 
we  find  that  equation  (2)  is  satisfied  by  the  root  3,  and  equa- 
tion (3)  by  the  root  —  ^.  The  other  two  equations  are 
impossible. 

Consequently,  in  solving  an  irrational  equation,  we  must 
expect  to  obtain  not  only  its  roots,  but  also  the  roots  of  the 
other  equations  obtained  by  changing  the  signs  of  the  radicals 
in  all  possible  ways.  Some  of  these  equations  will  be  impos- 
sible. The  roots  of  the  other  irrational  equations  will  be  the 
roots  of  the  rational  equation. 


292  ALGEBRA.  [Cn.  XVIII 

14.   Ex.     Solve  the  equation 

V(3  xz  -  2  x  +  4)  -  3  ar°  +  2  x  =  -  16. 
Since  -  3  y?  +  2  a?  =  -(3  or  -  2  a;  +  4)+  4, 

we  may  take  -^/(S  x2  —  2  a;  +  4)  as  the  unknown  number,  replac- 
ing it  temporarily  by  y.     We  then  have  the  quadratic  equation 

y  -  i/2  +  4  =  -  16. 

The  roots  of  this  equation  are  5,  and  —  4. 
Equating  ^/(3  y?  —  2  x  -f  4)  to  each  of  these  roots,  we  have 

-^(3  x2  —  2  x  +  4)  =  5,  whence  x  =  3,  —  J. 
V(3  y?  -  2  x  +  4)  =  -4,  whence  a;  =  J  (1  ±  V37). 
The  numbers  3,   —  J  satisfy  the  given  equation,  and  are 
therefore  roots  of  that  equation.     The  numbers  -§V(1  ±  V^7) 
do  not  satisfy  the  given  equation. 

But  if  the  value  of  the  radical  is  not  restricted  to  the  posi- 
tive root,  the  given  equation  comprises  the  two  equations 

V(3  «2  -  2  x  +  4)  -  3  a?  +  2  x  =  -  16,  (1) 

-V(3«2-2aj  +  4)-3oja  +  2a;  =  -16.  (2) 

Then  -|(1  ±^/37)  are  roots  of  (2). 
The  given  equation  is  said  to  be  in  quadratic  form. 

EXERCISES   VII. 

Solve  each  of  the  following  equations,  and  check  the  results. 
If  a  result  does  not  satisfy  an  equation  as  written,  determine 
what  signs  the  radical  terms  must  have  in  order  that  the 
result  may  satisfy  the  equation. 

2.   4z 
3.   3-3^-4^  +  9=0.      4.   5^ 


7 


14-15]  HIGHER  EQUATIONS.  293 


17 


25 


11.   r?^/x=3x2-\-3x-53.      12. 
13.   (5- 
15. 

16. 


20 

' 

21.   x2  -x  +  2V(^  -«-!!)  =  14. 

22. 

23. 

24. 


27. 


28. * __ 


HIGHER  EQUATIONS. 

15.   Certain  equations  of  higher  degree  than  the  second  can 
be  solved  by  means  of  quadratic  equations. 

Ex.  1.   Solve  the  equation      x3  —  1  =  0. 
Factoring,  (a;  —  1)  (x2  +  x  +  1)  =  0. 

This  equation  is  equivalent  to  the  two  equations 

x  —  1  =  0,  whence  x  =  1 ; 
and  x2  +  x  +  1  =  0,  whence  x  •=  —  £  ± 


294  ALGEBRA.  [Cn.  XVIII 

This  example  gives  the  three  cube  roots  of  1,  since  x3—  1=0 
is  equivalent  to 

x3  =  1,  or  x  =  -J-l. 

Therefore  the  three  cube  roots  of  1  are 


In  general,  the  three  cube  roots  of  any  number  can  be  found 
by  multiplying  the  arithmetical  cube  root  of  the  number  in 
turn  by  the  three  algebraic  cube  roots  of  1. 

^8  =  2^/1=2,  -l±V-3. 

Ex.  2.    Solve^the  equation  a4  -  9  =  2  x2  -  1. 
Since  x4  =  (x2)2,  we  may  take  x2  as  the  unknown  number  and 
solve  this  equation  as  a  quadratic  in  x2. 

We  then  have          (x2)2  -2x2-8,  =  0. 
Factoring,  (or2  -  4)  (x2  +  2)  =  0. 

Whence, 
x2  -  4  =  0,  or  x  =  ±  2  ;  and  x2  +  2  =  0,  or  a;  =  ±  V~  2. 

In  general,  any  equation  containing  only  two  powers  of  the 
unknown  number,  one  of  which  is  the  square  of  the  other,  can  be 
solved  as  a  quadratic  equation. 

Ex.  3.   Solve  the  equation  (x2—  3x+l)2=6+5(x2—  3a;+l). 
In  this  example  x2  —  3  x  -f-  1  is  regarded  as  the  unknown 
number,  and  may  temporarily  be  represented  by  the  letter  y. 
The  equation  then  becomes 

y2  =  6  +  5  y  ;  whence  y  =  6,  and  —  1. 
We  therefore  have  the  two  equations 

x2-3x  +  l  =  6,  whence  x  =  f  ±  iV295 
x2  —  3a?  +  l=  —  1,  whence  x  =  2,  a?  =  1. 

Therefore  the  roots  of  the  given  equation  are  f  ±  %^/2$,  2,  1. 

Attention  is  called  to  the  fact  that,  in  each  example,  we  have 
obtained  as  many  roots  as  there  are  units  in  the  degree  of  the 
equation. 


15-16]  PROBLEMS.  295 

EXERCISES  VIII. 

Solve  each  of  the  following  equations  : 

1.   3^  +  1  =  0.  2.   z4-l  =  0.  3.   a6  +  1=0. 

4.   a6-  1  =  0.  5.    (a;  -I)3  =  8.          6.   x?  =  (2a-x)3. 

7.    (^  +  1)4=16.      8.   «4  +  9  =  10rJ.       9.   x4-6x2  =  -l. 
10.   o;6-65arJ  =  -64.  11. 

12. 
13. 
14. 


q     a  -a.o          16 

' 


(a  +  »)8  +  (a  -  a?)8  a4  -  6ar>  +  1      2 

x2  —  a2     x2  -\-  a?  _  34: 


• 


'' 


PROBLEMS. 

16.  Pr.  1.  The  sum  of  two  numbers  is  15,  and  their  product 
is  56.  What  are  the  numbers  ? 

Let  x  stand  for  one  of  the  numbers  ;  then,  by  the  first  con- 
dition, 15  —  x  stands  for  the  other  number.  By  the  second 
condition 

x  (15  —  x)  =  56  ;  whence  x  =  7,  and  8. 

Therefore  x  =  7,  one  of  the  numbers,  and  15  —  x  =  8,  the 
other  number.  Observe-  that  if  we  take  x  =  8,  then  15  —  x  =  7. 
That  is,  the  two  required  numbers  are  the  two  roots  of  the 
quadratic  equation. 

Pr.  2.   Divide  100  into  two  parts  whose  product  is  2600. 
Let  x  stand  for  the  less  part,  and  100  —  x  for  the  greater. 
By  the  second  condition,  x  (100  -  x)  =  2600.     The  roots  of 
this  equation  are  50  +  10^/—  1  and  50  —  10  V~  1. 


296  ALGEBRA.  [Cn.  XVIII 

An  imaginary  result  always  indicates  inconsistent  conditions 
in  the  problem.  The  inconsistency  of  these  conditions  may  be 
shown  as  follows  : 

Let  d  stand  for  the  difference  between  the  two  parts  of  100. 
Then  50  -f-  £  d  stands  for  the  greater  part,  and  50  —  £  d  for  the 
less. 

The  product  of  the  two  parts  is 

(50  +  J  d)  (50  -  ±  d),  =  2500  -  Q  cZ)2  =  2500  -  J  d2. 

Since  d2  is  positive  for  all  real  values  of  d,  the  product  2500 
—  \  d2  must  be  less  than  2500.  Consequently  100  cannot  be 
divided  into  two  parts  whose  product  is  greater  than  2500. 

17.  When  the  solution  of  a  problem  leads  to  a  quadratic 
equation,  it  is  necessary  to  determine  whether  either  or  both 
of  the  roots  of  the  equation  satisfy  the  conditions  expressed 
and  implied  in  the  problem. 

Positive  results,  in  general,  satisfy  all  the  conditions  of  the 
problem. 

A  negative  result,  as  a  rule,  satisfies  the  conditions  of  the 
problem,  when  they  refer  to  abstract  numbers.  When  the 
required  numbers  refer  to  quantities  which  can  be  understood 
in  opposite  senses,  as  opposite  directions,  etc.,  an  intelligible 
meaning  can  usually  be  given  to  a  negative  result. 

An  imaginary  result  always  implies  inconsistent  conditions. 

18.  The  interpretation  of  a  negative  result  is  often  facili- 
tated by  the  following  principle: 

If  a  given  quadratic  equation  have  a  negative  root,  then  the 
equation  obtained  by  changing  the  sign  of  x  has  a  positive  root 
of  the  same  absolute  value. 

E.g.,  the  roots  of  the  equation  y?  —  5  #  +  6  =  0  are  2  and  3  ; 
and  the  roots  of  the  equation 


or  a^-f  5  x  +  6  =  0,  are  —  2  and  —  3. 


16-18]  PROBLEMS.  297 

Pr.  3.  A  man  bought  muslin  for  $  3.00.  If  he  had  bought 
3  yards  more  for  the  same  money,  each  yard  would  have  cost 
him  5  cents  less.  How  many  yards  did  he  buy  ? 

Let  x  stand  for  the  number  of  yards  the  man  bought.     Then 

800 
1  yard  cost  -  -  cents.     If  he  had  bought  x  +  3  yards  for  the 

x  300 

same  money,  each  yard  would  have  cost  -  - —  cents. 

^nn       ^nn  x  -\-  o 

Therefore  ^--^-=5;  whence  x  =  12  and  -15. 

x       x  +  3 

The  root  12  satisfies  the  equation  and  also  the  conditions  of 
the  problem  ;  the  root  —  15  has  no  meaning. 

But  if  x  be  replaced  by  —  x  in  the  equation,  we  obtain  a  new 
equation, 

300         300         -  300       300      K 

——  =  5,  or =  5,  (2) 

—  x      — ic  +  3  a;  — 3       x 

whose  roots  are  —  12  and  -f  15. 

Equation  (2)  evidently  corresponds  to  the  problem :  A  -man 
bought  muslin  for  $  3.00.  If  he  had  bought  3  yards  less  for 
the  same  money,  each  yard  would  have  cost  him  5  cents  more. 

Notice  that  the  intelligible  result,  12,  of  the  first  statement 
has  become  —  12  and  is  meaningless  in  the  second  statement. 

EXERCISES    IX. 

1.  If  1  be  added  to  the  square  of  a  number,  the  sum  will  be 
50.     What  is  the  number  ? 

2.  If  5  be  subtracted  from  a  number,  and  1  be  added  to  the 
square  of  the  remainder,  the  sum  will  be  10.     What  is  the 
number  ? 

3.  One  of  two  numbers  exceeds  50  by  as  much  as  the  other 
is  less  than  50,  and   their  product  is  2400.     What  are  the 
numbers  ? 

4.  The  product  of  two  consecutive   integers   exceeds  the 
smaller  by  17,424.     What  are  the  numbers  ? 

5.  If  27  be  divided  by  a  certain  number,  and  the  same  num- 
ber be  divided  by  3,  the  results  will  be  equal.     What  is  the 
number  ? 


298  ALGEBRA.  [Cn.  XVIII 

6.  What  number,  added  to  its  reciprocal,  gives  2.9  ? 

7.  What  number,  subtracted  from  its  reciprocal,  gives  n  ? 
Let  n  =  6.09. 

8.  If  n  be  divided  by  a  certain  number,  the  result  will  be 
the  same  as  if  the  number  were  subtracted  from  n.     What  is 
the  number  ?     Let  n  =  4. 

9.  If  the  product  of  two  numbers  be  176,  and  their  differ- 
ence be  5,  what  are  the  numbers  ? 

10.  A  certain  number  was  to  be  added  to  £,  but  by  mistake 
-J-  was  divided  by  the  number.     Nevertheless,  the  correct  result 
was  obtained.     What  was  the  number  ? 

11.  If  100  marbles  be  so  divided  among  a  certain  number  of 
boys  that  each  boy  shall  receive  four  times  as  many  marbles  as 
there  are  boys,  how  many  boys  are  there  ? 

12.  The  area  of  a  rectangle,  one  of  whose  sides  is  7  inches 
longer  than  the  other,  is  494  square  inches.     How  long  is  each 
side? 

13.  The  difference  between  the  squares  of  two  consecutive 
numbers  is  equal  to  three  times  the  square  of  the  less  number. 
What  are  the  numbers  ? 

14.  A  merchant  received  $  48  for  a  number  of  yards  of  cloth. 
If  the  number  of  dollars  a  yard  be  equal  to  three-sixteenths  of 
the  number  of  yards,  how  many  yards  did  he  sell  ? 

15.  In  a  company  of  14  persons,  men  and  women,  the  men 
spent  $  24  and  the  women  $  24.     If  each  man  spent  $  1  more 
than  each  woman,  how  many  men  and  how  many  women  were 
in  the  company  ? 

16.  A  pupil  was  to  add  a  certain  number  to  4,  then  to  sub- 
tract the  same  number  from  9,  and  finally   to  multiply  the 
results.     But  he  added  the  number  to  9,  then  subtracted  4  from 
the  number,  and   multiplied  these  results.     Nevertheless  he 
obtained  ,the  correct  product.     What  was  the  number  ? 

17.  A  man  paid  $  80  for  wine.     If  he  had  received  4  gallons 
less  for  the  same  money,  he  would  have  paid  $  1  more  a  gallon. 
How  many  gallons  did  he  buy  ? 


18]  PROBLEMS.  299 

18.  A  man  left  $  31,500  to  be  divided  equally  among  his 
children.     But  since  3  of  the  children  died,  each  remaining 
child  received  $  3375  more.     How  many  children  survived  ? 

19.  Two  bodies  move  from  the  vertex  of  a  right  angle  along 
its  sides  at  the  rate  of  12  feet  and  16  feet  a  second  respectively. 
After  how  many  seconds  will  they  be  90  feet  apart  ? 

20.  A  tank  can  be  filled  by  two  pipes,  by  the  one  in  two 
hours   less  time  than  by  the  other.      If  both  pipes  be  open 
1J  hours,  the  tank  will  be  filled.     How  long  does  it  take  each 
pipe  to  fill  the  tank  ? 

21.  From  a  thread,  whose  length  is  equal  to  the  perimeter 
of  a  square,  36  inches  are  cut  off,  and  the  remainder  is  equal 
in  length  to  the  perimeter  of  another  square  whose  area  is 
four-ninths  of  that  of  the  first.     What  is  the  length  of  the 
thread  ? 

22.  A  number  of  coins  can  be  arranged  in  a  square,  each 
side  containing  51  coins.     If  the  same  number  of  coins  be 
arranged  in  two  squares,  the  side  of  one  square  will  contain 
21  more  coins  than  the  side  of  the  other.     How  many  coins 
does  the  side  of  each  of  the  latter  squares  contain  ? 

23.  A  farmer  wished  to  receive  $  2.88  for  a  certain  number 
of  eggs.     But  he  broke  6  eggs,  and  in  order  to  receive  the 
desired  amount  he  increased  the  price  of  the  remaining  eggs 
by  2|-  cents  a  dozen.     How  many  eggs  had  he  originally  ? 

24.  Two  bodies  move  toward  each  other  from  A  and  B  re- 
spectively, and  meet  after  35  seconds.     If  it  takes  the  one  24 
seconds  longer  than  the  other  to  move  from  A  to  B,  how  long 
does  it  take  each  one  to  move  that  distance  ? 

25.  It  takes  a  boat's  crew  4  hours  and  12  minutes  to  row  12 
miles  down  a  river  with  the  current,  and  back  again  against 
the  current.     If  the  speed  of  the  current  be  3  miles  an  hour, 
at  what  rate  can  the  crew  row  in  still  water  ? 

26.  A  man  paid  $  300  for  a  drove  of  sheep.   By  selling  all  but 
10  of  them  at  a  profit  of  $  2.50  each,  he  received  the  amount 
he  paid  for  all  the  sheep.     How  many  sheep  did  he  buy  ? 


CHAPTER  XIX. 

SIMULTANEOUS   QUADRATIC   AND   HIGHER 
EQUATIONS. 

1.  The  solution  of  a  system  of  quadratic  or  higher  equa- 
tions in  general  involves  the  solution  of  an  equation  of  higher 
degree  than  the  second,  and  therefore  cannot  be  effected  by 
the  methods  for  solving  quadratic  equations.     But  there  are 
many  special  systems  whose  solutions  can  be  made  to  depend 
upon  the  solutions  of  quadratic  equations. 

The  proofs  of  the  following  methods  are  given  in  School 
Algebra,  Ch.  XXIV. 

2.  Elimination  by  Substitution.  —  When  one  equation  of  a 
system  of  two  equations  is  of  the  first  degree,  the  solution  can 
be  obtained  by  the  method  of  substitution. 

Ex.    Solve  the  system  y  +  2  x  =  5,      j  (1) 

rf-jf  =  -8.J  (2) 

Solving  (1)  for  y}  y  =  5  -  2  x.  (3) 

Substituting  5  —  2  x  for  y  in  (2), 


From  this  equation  we  obtain  x  =  1, 

and  x  =  5J. 

Substituting  1  for  x  in  (3),  y  =  3. 

Substituting  5|  for  x  in  (3),  y  =  —  6^-. 

It  is  proved  in  School  Algebra,  Ch.  XXIV.,  that  the  above 
method  is  based  upon  equivalent  equations. 

Therefore  the  solutions  of  the  given  system  are  1,  3  ;  5f  , 
—  6J,  the  first  number  of  each  pair  being  the  value  of  x,  and 
the  second  the  corresponding  value  of  y. 

300 


1-3]         SIMULTANEOUS   QUADRATIC    EQUATIONS.        301 

Had  we  substituted  1  for  x  in  (2),  we  should  have  obtained 
y  =  ±3. 

But  the  solution  1,  —  3  does  not  satisfy  equation  (1). 

Therefore,  always  substitute  in  the  linear  equation  the  value 
of  the  unknown  number  obtained  by  elimination. 

3.  Elimination  by  Addition  and  Subtraction.  —  This  method 
can  frequently  be  applied. 

Ex.    Solve  the  system  x2  +  3  y  =  18,  1  (1) 

2x2-5y  =  3.    J  (2) 

We  will  first  eliminate  y. 

Multiplying  (1)  by  5,  5  x2  +  15  y  =  90.  (3) 

Multiplying  (2)  by  3,  6  x?  -  15  y  =  9.  (4) 

Adding  (3)  and  (4),  11  x2  =  99. 

Whence,  x  =  3,  and  x  —  —  3. 

Substituting  3  for  x  in  (1),  y  =  3. 

Substituting  —  3  for  x  in  (1),  y  =  3. 

The  given  system  has  the  two  solutions  3,  3  ;  —  3,  3. 
Notice  that  this  example  could  also  have  been  solved  by  the 
method  of  substitution. 

EXERCISES  I. 

Solve  each  of  the  following  systems  : 


2.    <  3. 


4  , 

"   U2/  =  96.  '' 

(3x-2y  =  l, 
''  * 


+  6. 


302  ALGEBRA. 

Sx-J  +  tf-l,  r 

-L3.    1    _  ~  -  1*.    i 


[Cn.  XIX 


15'  {l°x-y  =  7+f  16'  { 


or2  +  5  xy  +  /  =  43, 


17. 


19. 


4      3 

*      / 


17 

7* 


x  —  y 


18. 


20. 


5a;-2»  ±  = 


4.  Homogeneous  Equations.  —  When  all  the  terms  which  con- 
tain the  unknown  numbers  in  both  equations  of  the  system  are 
of  the  second  degree,  a  system  can  always  be  derived  whose 
solution  is  obtained  by  the  method  of  Art.  2. 


Ex.    Solve  the  system 


Multiplying  (1)  by  73,  73  x2  +  73  xy  +  146  y2  =  74  x  73. 
Multiplying  (2)  by  74,  148  x2  +  148  xy  +  74  y2  =  74  x  73. 
Subtracting  (3)  from  (4),  75  x2  +  75  xy  -  72  y2  =  0, 


(1) 
(2) 
(3) 
(4) 


(5  x  -  3  y)  (5  x  +  8  y)  =  0. 
Therefore  the  given  system  is  equivalent  to 


a*  +  xy  + 


22/2  =  74,j(a)' 


x>  +  xy  +  2y*  = 


The  solutions  of  these  systems,  and  hence  of  the  given 
system,  are  respectively  3,  5 ;  —  3,  —  5 ;  8,  —  5 ;  —  8,  5. 

In  applying  this  method  to  such  systems,  we  must  first 
derive  from  the  given  equations  a  homogeneous  equation  in 
which  there  is  no  term  free  from  the  unknown  numbers. 


3-6]         SIMULTANEOUS   QUADRATIC   EQUATIONS.        303 

5.   Such  examples  can  also  be  solved  by  a  special  device. 

Ex.     Solve  the  system         x2  +  4  f  =  13,  (1) 

^  +  2^  =  5.  (2) 

In  both  equations,  let  y  =  tx.  (3) 

Then  from  (1),  y?  +  4  aft2  =  13,  whence  a8  =  ^   13,  9  ;          (4) 

1  +  4  r 

and  from  (2),          x2t  +  2tft2  =  5,  whence  y?  =  —  5__-  (5) 

Equating  values  of  x2,         13      =  —  -  --  (6) 

1  +  4*2     £  +  2£2 

Whence  t  =  1,  and  £  =  —  f  . 

When  «  =  i,  x-  =      13    9  =  9,  whence  a;  =  ±  3. 

When  t  =  —  %,  x2  =  1,  whence  a;  =  ±  y^- 
When  a;  =  ±  3,  y  =  tx  =  %(±  3)=±  1. 

When  a  =  ±  vi,  y  =  -  1(±  Vi)=  *  IVi- 

EXERCISES  II. 
Solve  each  of  the  following  systems  : 


f  aj  -  a 
'  '          -  2xy  +  15  =  0. 


'  ' 


g  =  61  -  3xy, 

'          -y2  =  31  -  2xy. 

6.  Symmetrical  Equations.  —  A  Symmetrical  Equation  is  one 
which  remains  the  same  when  the  unknown  numbers  are 
interchanged. 

A  system  of  two  symmetrical  equations  can  be  solved  by  first 
finding  the  values  of  x  -{-y  and  x  —  y. 


304                                         ALGEBRA.  [Cn.  XIX 

Ex.  1.    Solve  the  system               x2  +  y2  =  13,  j  (1) 

xy  =  G.    J  (2) 

Multiplying  (2)  by  2,                         2  o;y  =  12.  (3) 

Adding  (3)  to  (1),               x*  +  2  xy  +  f  =  25.  (4) 

Subtracting  (3)  from  (1),  x2  —  2xy  +  y2  =  l.  (5) 

Equating  square  roots  of  (4),          x  +  y  =  ±5.  (6) 

Equating  square  roots  of  (5),          x  —  y  =  ±  1.  (7) 

It  is  proved  in  School  Algebra,  Ch.  XXIV.,  that  equations 
(6)  and  (7)  are  equivalent  to 

a;  +  y  =  5,  *  +  #  =  5,       #  +  ?/  =  —  5,  1      a;  +  y  =  —  5, 

0  —    « 


*  +  #  =  5,  j    #  +  ?/  =  —  5,  1      a;  +  y  =  —  5, 
-y  =  -l,J     x-y  =  +l,)      a-2/  =  -l. 


The  solutions  of  these  four  systems  are  respectively  3,  2  ; 
2,3;  -2,  -3;  -3,  -2. 

The  solutions  of  (6)  .and  (7)  should  be  obtained  mentally, 
without  writing  the  equivalent  systems.  Each  sign  of  the 
second  member  of  (6)  should  be  taken  in  turn  with  each  sign 
of  the  second  member  of  (7). 

Notice  that  these  solutions  differ  only  in  having  the  values 
or  x  and  y  interchanged.  This  we  should  expect  from  the 
definition  of  symmetrical  equations. 

When  the  equations  are  symmetrical,  except  for  sign,  the 
solution  can  be  obtained  by  a  similar  method. 

Ex.  2.     Solve  the  system 

•r-f-4  (1) 

x2  +  f  =  29.  (2) 

Squaring  (1),  tf  -  2  xy  +  f  =  9,  (3) 

Subtracting  (3)  from  (2), 

2  xy  =  20,  or  xy  =  10.  (4) 

The  solutions  of  (1)  and  (4)  are  5,  2  ;   -  2,  -  5. 

Notice  that  the  solutions  in  this  case  differ  not  only  in  hav- 
ing the  values  of  x  and  y  interchanged,  but  also  in  sign. 


6]  SIMULTANEOUS   QUADRATIC   EQUATIONS.         305 


EXERCISES   III. 

Solve  each  of  the  following  systems : 


L    I  aw  =  32. 


xy  =  —  15. 


2. 


5. 


. 

xy  =  b. 


xy  =  28. 
3  xy  =  -  60. 


ay  =  12. 

io    |9x2  +  2/2  =  37a2, 


L  ay  =  -  90. 


r 

I  cc   =  2  a5. 


12.    \  •'     13. 

la?  +  y  =  15. 

(x2-y2  =  28,  f  x2-4:y2=  -3. 

15.    i  16.    j  17. 

I  x?/  =  48.  I  xy  =  —  1. 


x  —  y  =  5. 
16  ^+49  ?/2=113, 


(x  —  y  =  Z.              (6x—y  =  w. 

l 

f  a#  =  80, 

a?  +  y  =  16, 

21.       1       1  _  1         22. 

1      1_1 

23. 

24.                                 25.    ^ 

'a?     y      16 
y      a?  ~  15' 

26. 

—  =2. 

x-y  =  2. 

f  x  +  ccw  +  v  =  29,                         f 

27.                           9                            28. 

^  +  ?y2- 

f  x2  H-  w2  -  (aj  -  v)  =  20,                f 
29.                                                          30.    ^ 

*+f 

?/- 


- 


„. 


=  6. 


32 

" 


+  /-a^  =  21. 


306  ALGEBRA.  [Cn.  XIX 

7.  Higher  Equations.  —  The  solutions  of  certain  equations  of 
higher  degree  than  the  second  can  be  made  to  depend  upon  the 
solutions  of  quadratic  equations. 

Ex.  l.   Solve  the  system  x?  +  ys  =  9,  (1) 

a?  +  y  =  3.  (2) 

Dividing  (1)  by  (2),    x2— xy+y2  =  3.  (3) 
Subtracting  (3)  from  the  square  of  (2), 

3  xy  =  6,  or  xy  =  2.  (4) 
The  solutions  of  (2)  and  (4),  and  therefore  of  the  given 
system,  are  1,  2,  and  2,  1. 

Ex.  2.   Solve  the  system  x4  +  f  =  17,  (1) 

x  +  y  =  3.  (2) 
We  first  find  the  value  of  xy. 

Let                                              xy  =  z.  (3) 

•Squaring  (2),          x2  +  2  xy  +  y2  =  9,  (4) 

or                                               cu2  +  2/2  =  9  —  2  z.  (5) 

Squaring  (5),        x4  +  2  x2y2  +  i/4  =  81  -  36  z  +  4  z2,  (6) 

.Since  #4  +  y4  =  17,  we  have  from  (7), 

2z2-  36z  +  81  =  17.  (8) 

Whence                                       z  =  16,  and  2.  (9) 

Therefore,  from  (3)  and  (9),   xy  =  16,  (10) 

and    .                                                xy  =  2.  (11) 

The  solutions  of  (2)  and  (10)  and  of  (2)  and  (11)  are  readily 
found,  and  should  be  checked  by  substitution. 

EXERCISES   IV. 

Solve  each  of  the  following  systems : 

2'    (xs-y*  =  7. 

-  2)3=  28, 

i    ^  4.    '  x          '    '  ^ 

'    l32(x3  +  2/3)  =  2285. 


7-8]         SIMULTANEOUS   QUADRATIC   EQUATIONS.        307 


Problems. 

8.  Pr.  The  front  wheel  of  a  carriage  makes  6  more  revolu- 
tions than  the  hind  wheel  in  travelling  360  feet.  But  if  the 
circumference  of  each  wheel  were  3  feet  greater,  the  front 
wheel  would  make  only  4  revolutions  more  than  the  hind 
wheel  in  travelling  the  same  distance  as  before.  What  are 
the  circumferences  of  the  two  wheels  ? 

Let  x  stand  for  the  number  of  feet  in  the  circumference  of 
front  wheel,  and  y  for  the  number  of  feet  in  the  circumference  of 
hind  wheel.  Then  in  travelling  360  feet  the  front  wheel  makes 

360  360 

revolutions,  and  the  hind  wheel  makes revolutions. 

x  y 

By  the  first  condition,  —  =  —  +  6.  (1) 

x         y 

If  3  feet  were  added  to  the  circumference  of  each  wheel,  the 

front  wheel  would  make  —     —  revolutions,  and  the  hind  wheel 

360  x  -\-  o 
revolutions. 


By  the  second  condition,          -  =  -         +  4.  (2) 

x+3     y +o 

Whence  x  =  12,  the  circumference  of  the  front  wheel,  and 
y  =  15,  the  circumference  of  the  hind  wheel. 


308  ALGEBRA.  [Cn.  XIX 


EXERCISES   V. 

1.  The   square   of   one   number   increased  by  ten  times  a 
second  number  is  84,  and  is  equal  to  the  square  of  the  second 
number  increased  by  ten  times  the  first. 

2.  The  sum  of  two  numbers  is  20,  and  the  sum  of  the  square 
of  the  one  diminished  by  13  and  the  square  of  the  other  in- 
creased by  13  is  272.     What  are  the  numbers  ? 

3.  Find  two  numbers  such  that  their  difference  added  to 
the  difference  of  their  squares  shall  be  150,  and  their  sum 
added  to  the  sum  of  their  squares  shall  be  330. 

4.  Find  two  numbers  whose  sum  is  equal  to  their  product 
and  also  to  the  difference  of  their  squares. 

5.  The  sum  of  the  fourth  powers  of  two  numbers  is  1921, 
and  the  sum  of  their  squares  is  61.     What  are  the  numbers  ? 

6.  If  a  number  of  two  digits  be  multiplied  by  its  tens'  digit, 
the  product  will  be  390.     If  the  digits  be  interchanged  and  the 
resulting  number  be  multiplied  by  its  tens'  digit,  the  product 
will  be  280.     What  is  the  number  ? 

7.  If  a  number  of  two  digits  be  divided  by  the  product  of 
its  digits,  the  quotient  will  be  2.     If  27  be  added  to  the  number, 
the  sum  will  be  equal  to  the  number  obtained  by  interchanging 
the  digits.     What  is  the  number  ? 

8.  The  product  of  the  two  digits  of  a  number  is  equal  to 
one-half  of  the  number.     If  the  number  be  subtracted  from 
the  number  obtained  by  interchanging  the  digits,  the  remainder 
will  be  equal  to  three-halves  of  the  product  of  the  digits  of 
the  number.     What  is  the  number  ? 

9.  If  the    difference  of   the   squares   of  two  numbers   be 
divided  by  the  first  number,  the  quotient  and  the  remainder 
will  each  be  5.     If  the  difference  of  the  squares  be  divided 
by  the  second  number,  the  quotient  will  be  13  and  the  re- 
mainder 1.     What  are  the  numbers  ? 


8]  SIMULTANEOUS   QUADRATIC   EQUATIONS.         309 

10.  The  sum  of  the  three  digits  of  a  number  is  9.     If  the 
digits  be  written  in  reverse  order,  the  resulting  number  wil} 
exceed  the  original  number  by  396.     The  square  of  the  middle 
digit  exceeds  the  product  of  the  first  and  third  digit  by  4. 
What  is  the  number  ? 

11.  A  rectangular  field  is  119  yards  long  and  19  yards  wide. 
How  many  yards  must  be  added  to  its  width  and  how  many 
yards  must  be  taken  from  its  length,  in  order  that  its  area  may 
remain  the  same  while  its  perimeter  is  increased  by  24  yards  ? 

12.  The   floor   of   a  room  contains  30J  square  yards;  one 
wall  contains  21  square  yards,  and  an  adjacent  wall  contains 
13  square  yards.     What  are  the  dimensions  of  the  room  ? 

13.  A.  merchant  bought  a  number  of  pieces  of  cloth  of  two 
different  kinds.     He  bought  of  each  kind  as  many  pieces  and 
paid  for  each  yard  half  as  many  dollars  as  that  kind  contained 
yards.     He  bought  altogether  19  pieces  and  paid  for  them 
$  921.50.     How  many  pieces  of  each  kind  did  he  buy  ? 

14.  The  diagonal  of  a  rectangle  is  20|  feet.     If  the  lengtk 
of  one  side  be  increased  by  14  feet  and  the  length  of  the  other 
side  be  diminished  by  2f  feet,  the  diagonal  will  be  increased 
by  124  feet.    What  are  the  lengths  of  the  sides  of  the  rectangle  ? 

15.  A  certain  number  of  coins  can  be  arranged  in  the  form 
of  one  square,  and  also  in  the  form  of  two  squares.     In  the 
first  arrangement  each  side  of  the  square  contains  29  coins, 
and  in  the  second  arrangement  one  square  contains  41  more 
coins  than  the  other.     How  many  coins  are  there  in  a  side  of 
each  square  of  the  second  arrangement  ? 

16.  A  piece  of  cloth  after  being  wet  shrinks  in  length  by 
one-eighth  and  in  breadth  by  one-sixteenth.     The  piece  con- 
tains  after   shrinking   3.68  fewer   square   yards  than  before 
shrinking,  and  the   length  and  breadth  together  shrink   1.7 
yards.     What  was  the  length  and  breadth  of  the  piece  ? 

17.  A  merchant  paid  $  125  for  two  kinds  of    goods.     He 
sold  the  one  kind  for  $  91  and  the  other  for  $  36.    He  thereby 


310  ALGEBRA.  [Cn.  XIX 

gained  as  much  per  cent  on  the  first  kind  as  ho.  lost  on  the 
second.     How  much  did  he  pay  for  each  kind  ? 

18.  Two  workmen  can  do  a  piece  of  work  in  6  days.     How 
long  will  it  take  each  of  them  to  do  the  work,  if  it  takes  one 
5  days  longer  than  the  other  ? 

19.  Two  men,  A  and  B,  receive  different  wages.     A  earns 
$  42,  and  B  $  40.     If  A  had  received  B's  wages  a  day,  and  B 
had  received  A's  wages,  they  would  have  earned  together  $  4 
more.     How  many  days  does  each  work,  if  A  works  8  days 
more  than  B,  and  what  wages  does  each  receive  ? 

20.  It  takes  a  number  of  workmen  8  hours  to  remove  a  pile 
of  stones  from  one  place  to  another.     Had  there  been  8  more 
•workmen,  and  had  each  one  carried  5  pounds  less  at  each  trip, 
they  would  have  completed  the  work  in  7  hours.     Had  there 
been  8  fewer  workmen  and  had  each  one  carried  11  pounds 
more  at  each  trip,  they  would  have  completed  the  work  in  9 
hours.     How  many  workmen  were  there  and  how  many  pounds 
-did  each  one  carry  at  every  trip  ? 

21.  A  tank  can  be  filled  by  one  pipe  and  emptied  by  another. 
If,  when  the  tank  is  half  full  of  water,  both  pipes  be  left  open 
12  hours,  the  tank  will  be  emptied.     If  the  pipes  be  made 
smaller,  so  that  it  will  take  the  one  p'ipe  one  hour  longer  to 
fill  the  tank  and  the  other  one  hour  longer  to  empty  it,  the 
tank,  when  half  full  of  water,  will  then  be  emptied  in   15} 
Iiours.     In  what  time  will  the  empty  tank  be  filled  by  the  one 
rpipe,  and  the  full  tank  be  emptied  by  the  other  ? 


CHAPTER   XX. 

RATIO,   PROPORTION,    AND   VARIATION. 

RATIO. 

1.  The  Ratio  of  one  number  to  another  is  the  relation  between; 
the  numbers  which  is  expressed  by  the  quotient  of  the  first, 
divided  by  the  second. 

E.g.,  the  ratio  of  6  to  4  is  expressed  by  f ,  =  f . 

The  ratio  of  one  number  to  another  is  frequently  expressed 
by  placing  a  colon  between  them  ;  as  5  :  7. 

The  first  number  in  a  ratio  is  called  the  First  Term,  or  the 
Antecedent  of  the  ratio,  and  the  second  number  the  Second 
Term,  or  the  Consequent  of  the  ratio. 

Thus,  in  the  ratio  a :  b,  a  is  the  first  term,  and  b  the  second. 

2.  Since,  by  definition,  a  ratio  is  a  fraction,  all  the  proper- 
ties of  fractions  are  true  of  ratios ;  as  a  :  b  =  ma  :  mb. 

3.  The  definition  given  in  Art.  1  has  reference  to  the  ratio 
of  one  number  to  another.      But  it  is  frequently  necessary  to 
compare  concrete  quantities,  as  the  length  of  one  line  with  the 
length  of  another  line,  etc. 

If  two  concrete  quantities  of  the  same  kind  can  be  expressed  by 
two  rational  numbers  in  terms  of  the  same  unit,  then  the  ratio  of 
the  one  quantity  to  the  other  is  defined  as  the  ratio  of  the  one- 
number  to  the  other. 

2i     35 

E.g.,  the  ratio  of  2  L  yards  to  11  yards  is  2-J-:  1^,  =^-  =  — . 

1^-     16 

Observe  that  by  this  definition  the  ratio  of  two  concrete 
quantities  is  a  number.  Also  that  the  quantities  to  be  com- 
pared must  be  of  the  same  kind.  Dollars  cannot  be  compared, 
with  pounds,  etc. 

311 


312  ALGEBRA.  [Cn.  XX 

4.  If  two  concrete  quantities  cannot  be  expressed  by  two 
rational  numbers,  integers  or  fractions,  in  terms  of  the  same 
unit,  they  are  said  to  be  Incommensurable  one  to  the  other. 

Thus,  if  the  lengths  of  the  two  sides  of  a  right  triangle  be 
equal,  the  length  of  the  hypothenuse  cannot  be  expressed  by  a 
rational  number  in  terms  of  a  side  as  a  unit,  or  any  fraction  of 
a  side  as  a  unit. 

If  a  side  be  taken  as  the  unit,  the  hypothenuse  is  expressed 
by  -Y/2,  an  irrational  number.  And  the  ratio  of  the  hypothe- 
nuse to  a  side  is  >/2  :  1,  =  -^/2.  But  as  was  shown  in  Ch.  XV, 
Art.  40,  an  approximate  value  of  -y/2  can  be  found  to  any 
required  degree  of  accuracy. 

5.  In  general  let  P  and  Q  be  two  incommensurable  quan- 
tities.    It  is  proved  in  School  Algebra,  Ch.  XXV,  that  two 

rational  numbers  —  and  m     —  can  be  found,  between  which 
n  n 

the  value  of  the  ratio  P :  Q  lies.    These  two  fractions  differ  by 

-.     Therefore,  the  ratio  P:  Q,  which  lies  between  them,  differs 
n  .j 

from  either  of  them  by  less  than  -.     By  taking  n  sufficiently 

:great  we  can  make  -  as  small  as  we  please,  that  is,  less  than 
n 

<tny  assigned  number,  hoivever  small. 

It  is  also  proved  that  the  ratio  of  two  incommensurable 
quantities  is  a  number  which  obeys  the  fundamental  laws  of 
.algebra. 

It  is  therefore  not  necessary,  in  the  principles  of  this  chapter, 
to  make  any  distinction  between  such  ratios  and  those  which 
•can  be  expressed  exactly  in  terms  of  integers  and  fractions. 

EXERCISES  I. 
What  is  the  ratio  of 

1.   6  a  to  9  b  ?         2.   |  a2b  to  T6T  aW  ?       3.   9J  x*y  to  7|  xy*  ? 

4-H? 


4-8]  PROPORTION.  313 

7.   Which  is  the  greater  ratio, 

a  +  2b:a  +  b  or  a  +  3b:a  +  2b? 
What  is  the  value  of  the  ratio  x  :  y 


a  if          ==io?  9.  if          =5? 

3x  —  y  3x-2y 

If  the  value  of  the  ratio  x  :  y  is  f  ,  what  is  the  value 

10.  ofWx~y?  11.  of  5£±«2? 

15x  +  y  3x-2y 

PROPORTION. 

6.  A   Proportion   is   an   equation  whose   members   are  two 
equal  ratios. 

E.g.,  4:3  =  8:6,  read  the  ratio  of  4  to  3  is  equal  to  the  ratio 
of  8  to  6,  or  4  is  to  3  as  8  is  to  6. 

Instead  of  the  equality  sign  a  double  colon  is  frequently 
used  ;  as  4  :  3  :  :  8  :  6. 

7.  Four  numbers  are  said  to  be  in  proportion,  or  to  be  pro- 
portional, when  the  first  is  to  the  second  as  the  third  is  to  the 
fourth. 

E.g.,  the  numbers  4,  3,  8,  6  are  proportional,  since  4:3  =  8:6. 

The  individual  numbers  are  called  the  Proportionals,  or 
Terms  of  the  proportion. 

The  Extremes  of  a  proportion  are  its  first  and  last  terms  ;  as 
4  and  6  above. 

The  Means  of  a  proportion  are  its  second  and  third  terms  ; 
as  3  and  8  above. 

The  Antecedents  and  Consequents  of  a  proportion  are  the 
antecedents  and  consequents  of  its  two  ratios. 

E.g.,  4  and  8  are  the  antecedents,  and  3  and  6  the  conse- 
quents of  the  proportion  4:3  =  8:6. 

Principles  of  Proportions. 

8.  In  any  proportion  the  product  of  the  extremes  is  equal  to 
the  product  of  the  means. 


314  ALGEBRA.  [Cn.  XX 

If  a  :  b  =  c 
By  Art.  1, 


If  a  :  b  =  c  :  d,  we  are  to  prove  ad  =  be. 

a_c_ 
b~d 

Clearing  of  fractions,       ad  =  be. 


9.   If  the  product  of  two  numbers  be  equal  to  the  product  of 
two  other  numbers,  the  four  numbers  are  in  proportion. 

Let  ad  =  be. 

Dividing  by  bd,  |  =  |,  or  a:  b  =  c:  d-}  (1) 

o     d 

bycd,  -  =  -,  or  a:c  =  6:d;  (2) 


C        £ 


by  aft,  =  ^,  or  d:b  =  c:a;  (3) 

by  ac,  -  =  -,  or  d:c  =  b:a. 

c     a 

Interchanging  the  ratios  in  (1),  (2),  (3),  (4), 


by  ac,  -  =  -,  or  d:c  =  b:a.  (4) 

c     a 


c:d  =  a:b',  (5) 

b:d  =  a:c-)  (6) 

c  :  a  =  d  :  6  ;  (7) 

6  :  a  =  d  :  c.  (8) 

Notice  that  the  two  numbers  of  either  product  may  be  taken 
as  the  extremes,  the  other  two  as  the  means.  In  (1)  to  (4), 
o  and  d  are  the  extremes,  c  and  b  the  means  ;  in  (5)  to  (8), 
d  and  a  are  the  means,  c  and  b  the  extremes. 

10.  In  Art.  9,  we  may  regard  the  proportions  (2)  to  (8)  as 
being  derived  from  (1),  and  thus  obtain  the  following  proper- 
ties of  a  proportion  : 

(i.)   TJie  means  may  be  interchanged;  as  in  (2). 
(ii.)   The  extremes  may  be  interchanged;  as  in  (3). 
(iii.)   The  means  may  be  interchanged,  and  at  the  same  time 
the  extremes;  as  in  (4). 


8-13]  PROPORTION.  315 

(iv.)   Tlie  means  may  be  taken  as  the  extremes,  and  the  ex- 
tremes as  the  means;  as  (8)  from  (1),  (7)  from  (2),  etc. 

11.  If  any  three  terms  of  a  proportion  be  given,  the  remaining 
term  can  be  found. 

Ex.   What  is  the  second  term  of  a  proportion,  whose  first, 
third,  and  fourth  terms  are  10,  16,  and  8  respectively  ? 
Letting  x  stand  for  the  second  term,  we  have 

10:  a;  =  16:  8,  or  16a?  =  80;  whence  x  =  5. 

12.  The  products,  or  the  quotients,  of  the  corresponding  terms 
of  two  proportions  form  again  a  proportion. 

If  a:b  =  c:d,  or  2  =  £,  (1) 

u     a 

and  x  :  y  =  z  :  u,  or  -  =  -,  (2) 

we  have,  multiplying  corresponding  members  of  (1)  and  (2), 

—  =  —  ;  whence  ax  :  by  =  cz  :  du. 
by     du 

Dividing  the  members  of  (1)  by  the  corresponding  members 
of  (2),  we  have 

a     c 

x     z        ,  abed 

-  =  -  :  whence  -:-  =  -:  — 
b      d  x    y     z    u 

y     u 

13.  In  any  proportion,  the  sum  of  the  first  two  terms  is  to  the 
first  (or  the  second)  term  as  the  sum  of  the  last  two  terms  is  to 
the  third  (or  the  fourth)  term. 

Let  a  :  b  =  c  :  d. 

Then  «  =  L 

b     d 

Adding  1  to  both  members,  -  +  !  =  -  +  !, 

b  d 

or  a  +  b  _c  +  d 


316  ALGEBRA.  [Cn.  XX 

Whence  a  +  b:  b  =  c  +  did. 

In  like  manner  it  can  be  proved  that 
a  +  b  :  a  =  c  -f-  d  :  c. 

These  two  proportions  are  said  to  be  derived  fiom  the  given 
proportion  by  Composition. 

14.  In  any  proportion,  the  difference  of  the  first  two  terms  is  to 
the  first  (or  the  second)  term  as  the  difference  of  the  last -two 
terms  is  to  the  third  (or  the  fourth)  term. 

If  a:b  =  c:d, 

then         a  —  b:  a  =  c—  d:c,  and  a—b:b  =  c  —  d:d. 

The  proof  is  similar  to  that  of  Art.  13. 

These  two  proportions  are  said  to  be  derived  from  the  given 
proportion  by  Division. 

15.  In  any  proportion,  the  sum  of  the  first  two  terms  is  to 
their  difference  as  the  sum  of  the  last  two  terms  is  to  their 
difference. 

Let  a  :  b  =  c  :  d. 

By  Art.  13,  a-}-b:b  =  c  +  d:  d-, 

and  by  Art.  14,  a  —  b  :  b  =  c  —  d  :  d. 

mi        i  -10      a  -{-  b    1      c  -f-  d   1 

Then  by  Art.  12,  -  :  1  =  —    -  :  1, 

a  —  b  c  —  d 

a  -f-  b  _c  -f-  d 
a  —  b     c  —  d 

Whence  a  +  b:  a  —  b  =  c-\- d:  c  —  d. 

This  proportion  is  said  to  be  derived  from  the  given  one  by 
Composition  and  Division. 

16.  A  Continued  Proportion  is  one  in  which  the  consequent 
of  each  ratio  is  the  antecedent  of  the  following  ratio ;  as, 

a:b  =  b  :  c  =  c  :  d  =  etc. 


13-20]  PROPORTION.  317 

17.  In  the  continued  proportion 

a  :  b  =  b  :  c, 

h  is  called  a  Mean  Proportional  between  a  and  c,  and  c  is  called 
the  Third  Proportional  to  a  and  b. 

18.  The  mean  proportional  between  any  two  numbers  is  equal 
to  the  square  root  of  their  product 

From  a  :  b  =  b  :  c, 

we  have,  by  Art.  8,  W  =  ac  ;  whence  b  =  ^/(ac). 

19.  In  a  series  of  equal  ratios,  any  antecedent  is  to  its  conse- 
quent as  the  sum  of  all  the  antecedents  is  to  the  sum  of  all  the 
consequents. 

Let  ni  :  dl  =  n2  :  d2  =  n3  :  d3  =  •••  =  v, 

<h=v=v=v.... 


Then,  nl  =  vdl,  n2  =  vd2,  n3  =  vd3,  •  ••. 

Adding  corresponding  members  of  these  equations,  we  have 
ni  +  n2  +  n3  +  •  ••  =  vd^  +  vd2 


Therefore   ni  +  ^  +  ^3  +  '"  =  v  =^  =  ^= 
^1  +  ^2+^3+  •••  ch  .  d2 

1      4  +  5      10 


Ea 


2     8     10     2  +  8  +  10     20 


20.   The  following  examples  are  applications  of  the  preced- 
ing theory  : 

Ex.  1.   Find  a  mean  proportional  between  5  and  20. 
Let  x  stand  for  the  required  proportional. 

Then,  by  Art.  18,   x  =  V(5  X  20)  =  10. 
Ex.  2.    If  a  :  b  =  c  :  d, 

then  ab  +  cd  :  ab  -  cd  =  b2  +  d2  :  b2  -  d2. 

Let  *  =  c-  =  x. 

b     d 


318  ALGEBRA.  [CH.  XX 

Then  a  =  bx  and  c  =  dx. 

Therefore  ab  +  cd  =  Irx  -f  d2x, 

and  ab  —  cd  =  Irx  —  d2x. 


Whence      db  +  cd  :  ab  -  cd  =  b2  +  d2  :  b2  -  d2. 
Ex.  3.    Solve  the  equation 


By  composition  and  division, 


Squaring  and  clearing  of  fractions, 

2  +  x  =  18  —  9  x  ;  whence  a;  =  f  . 

EXERCISES   II. 

Verify  each  of  the  following  proportions  : 
1.    2i:li  =  li:J.  2.    14J  :  4|  =  200  :  60. 

4a&    tq2  +  &2=    2a6  1 

*   a2_&2:a_6      a4-64'2a-26' 

Form  proportions  from  each  of  the  following  products,  in 
eight  different  ways  : 

4.   2x  =  3y.  5.   m2  =  n2.  6.   a3  -  b3  =  x2  -  f. 

Find  a  fourth  proportional  to 
7.   1,  2,  and  8.         8.   -f,  f  ,  and  |.  9.   a&,  ac,  and  &. 

Find  a  third  proportional  to 

10.   2  and  6.  11.   \  and  \.  12.   a  and  b. 

Find  a  mean  proportional  between 

13.   2  and  18.          14.   \  and  f  .  15.   a?b  and  a&l 


20]  PROPORTION.  319 

Find  the  value  of  x  to  satisfy  each  of  the  following  pro- 
portions : 

18.    x:  2  =  12:3.       19.   161:253  =  8:407.       20. 
21. 


a  —  6 

Solve  each  of  the  following  equations  : 
23 


a-®)     V6 


26 


•^/(ax)  —  ba  —  b  a  +  b  a  —  b 

27    5a?  +  6  -2-  2  - 

' 


—  9 


= 

' 


3a;-2  4o;-3  6a;-7 

31.  Find  two  numbers  whose  ratio  is  7  :  5,  and  the  difference 
of  whose  square  is  96. 

32.  A  works  6  days  with  2  horses,  and  B  works  5  days  with 
3  horses.     What  is  the  ratio  of  A's  work  to  B's  work  ? 

33.  The  ratio  of  a  father's  age  to  his  son's  age  is  9:5.     If 
the  father  is  28  years  older  than  the  son,  how  old  is  each  ? 

34.  Find  three  numbers  in  a  continued  proportion  whose 
sum  is  39,  and  whose  product  is  729. 

35.  Find  two  numbers  such  that  if  one  be  added  to  the  first 
and  8  to  the  second,  the  sums  will  be  in  the  ratio  1  :  2,  and  if 
1  be  subtracted  from  each  number,  the  remainders  will  be  in 
the  ratio  2  :  3. 

36.  What  is  the  ratio  of  the  numerator  of  a  fraction  to  its 
denominator,  if  the  fraction  be  unchanged  when  a  is  added  to 
its  numerator  and  b  to  its  denominator  ? 

37.  The  sum  of  the  means  of  a  proportion  is  7,  the  sum  of 
the  extremes  is  8,  and  the  sum  of  the  squares  of  all  the  terms 
is  65.     What  is  the  proportion  ? 


3:20  ALGEBRA.  [Cn.  XX 

If  a:b  =  c:d,  prove  that 

38.  a  +  c:&  +  d  =  a2d:62c. 

39.  a2  +  &2:a2-62  =  c2  +  d2:cz-d*. 

40.  (a±b)2:ab  =  (c±d)2:cd. 

41.  2a  +  3&:4a  +  5&  =  2c  +  3d:4c  +  5d. 

42.  a  +  6  :  c  +  d  =  VO2  +  &2)  :  V(°2  +  d2)- 

43.  «2  +  &2  :       c2  +  ^2  =    73  +  &3  :  -/c3  +  d3  =  a  :  c. 


VARIATION. 

21.  Frequently  two  numbers  or  quantities  are  so  related  to 
each  other  that  a  change  in  the  value  of  one  produces  a  corre- 
sponding change  in  the  value  of  the  other. 

Thus,  the  distance  a  train  runs  in  one  hour  depends  upon 
its  speed,  and  increases  or  decreases  when  its  speed  increases 
or  decreases. 

The  illumination  made  by  a  light  depends  upon  the  intensity 
of  the  light,  and  varies  when  the  intensity  varies. 

The  value  of  y  given  by  the  equation  y  =  2x  —  3  depends 
upon  the  value  of  x,  and  varies  when  the  value  of  x  varies. 

Thus,  if  x  =  1,  y  =  —  1  ;  if  x  =  2,  y  =  l,  etc. 

We  shall  in  this  chapter  consider  only  the  simplest  kinds  of 
variation. 

22.  Direct  Variation.  —  T  wo  quantities  are  said  to  vary  directly 
one  as  the  other,  when  their  ratio  is  constant. 

Thus,  if  x  varies  directly  as  y,  then  -  =  &,  a  constant. 

*s 

For  example,  if  a  train  runs  at  a  uniform  speed,  the  number 
of  miles  it  runs  varies  directly  as  the  number  of  hours.  If  it 
runs  at  the  rate  of  30  miles  an  hour,  in  1  hour  it  will  run  30 
miles,  in  2  hours  60  miles,  in  3  hours  90  miles,  and  so  on  ;  and 
the  ratios  1  :  30,  2  :  60,  3  :  90,  etc.,  are  equal. 

The  symbol  of  direct  variation,  oc,  is  read  varies  directly  as. 

The  word  directly  is  frequently  omitted. 

If  y  =  3  x,  then  y  oc  x  (read  y  varies  as  x),  since  ^  =  3,  a 
constant. 


20-26]  VARIATION.  321 

23.  Inverse  Variation.  —  One  quantity  is  said  to  vary  inversely 
as  another  when  the  first  varies  as  the  reciprocal  of  the  second. 

Thus,  if  x  varies  inversely  as  y,  then  x  oc — 

y 

Therefore,  y  =  Jc,  a  constant ;  whence  xy  =  k. 

y 

That  is,  if  one  quantity  varies  inversely  as  another,  the 
product  of  the  quantities  is  constant. 

If  6  men  can  do  a  piece  of  work  in  12  hours,  3  men  can  do 
the  same  work  in  24  hours,  and  1  man  in  72  hours,  and  the 
products  6  x  12,  3  x  24,  1  x  72  are  equal.  That  is,  the  num- 
ber of  hours  varies  inversely  as  the  number  of  men  working. 
g 

If  y  =  -,  y  varies  inversely  as  x,  since  xy  =  3. 

»jC 

24.  Joint  Variation.  —  One  quantity  is  said  to  vary  as  two 
others  jointly,  when  it  varies  as  the  product  of  the  others. 

Thus,  if  x  varies  as  y  and  z  jointly,  then  —  =  Jc,  a  constant. 

yz 

For  example,  the  number  of  miles  a  train  runs  varies  as 
the  number  of  hours  and  the  number  of  miles  it  runs  an  hour 
jointly.  It  will  run  40  miles  in  2  hours  at  a  rate  of  20  miles 
an  hour,  90  miles  in  3  hours  at  the  rate  of  30  miles  an  hour, 

40  90  120 


and 


2  x  20     3  x  30     5  x  24 


25.  One  quantity  is  said  to  vary  directly  as  a  second  and 
inversely  as  a  third,  when  it  varies  as  the  second  and  the  recip- 
rocal of  the  third  jointly.  • 

Thus,  if  x  varies  directly  as  y  and  inversely  as  z,  then 

x  xz 

— -  =  K,  a  constant :  or  —  =  k. 
1  y 

y'z 

26.  In  all  the  preceding  cases  of  variation,  the  constant  can 
be  determined  when  any  set  of  corresponding  values  of  the 
quantities  is  known. 


322  ALGEBRA.  [Cn.  XX 

Ex.  1.    If  xccy,  and  x  =  3  when  y  =  5,  what  is  the  value  of 
the  constant  ? 

We  have  -  =  k.  or  x  =  ky. 

y 

Therefore,  when    x  =  3  and  y  =  5, 

3  =  5  fc,  whence  k  =  f . 
Consequently         #  =  f  ?/. 

EXERCISES   III. 

1.  If  a?  oc  i/,  and  x  =  10  when  y  =  5,  what  is  the  value  of  a? 
when  y  =  121  ? 

2.  If  x  oc  ?/,  and  x  =  a  when  y  =  J-  a2,  what  is  the  value  of  ?/ 
when  x  =  a26  ? 

3.  If  35  oc  2/2,  and  a?  =  5  when  y  =  —  3,  what  is  the  value  of  # 
when  y  =  15  ? 

4.  If  x  oc  -yA/,  and  x  =  a  +  m  when  y  =(a  —  m)2,  what  is 
the  value  of  a;  when  y  =  (a  +  w)4  ? 

5.  If  x  oc  — ,  and  x  =  3  when  y  =  J,  what  is  the  value  of  x 

«y 
when  y  =  4^  ? 

6.  If  #  oc  ^,  and  x  =  4  when  y  =  6  and  2  =  3,  what  is  the 

value  of  x  when  y  =  5,  and  2  =  2  ? 

7.  The  circumference  of  a  circle  whose  radius  is  6  feet  is 
37.7  feet.     What  is  the  circumference  of  a  circle  whose  radius 
is  9.5  feet,  if  it  be  known  that  the  circumference  varies  as  the 
radius  ? 

8.  An  ox  is  tied  by  a  rope  20  yards  long  in  the  centre  of  a 
field,  and  eats  all  the  grass  within  his  reach  in  2|  days.     How 
many  days  would  it  have  taken  the  ox  to  eat  all  the  grass 
within  his  reach  if  the  rope  had  been  10  yards  longer  ? 

9.  The  volume  of  a  sphere  whose  radius  is  7  inches  is 
1437.3  cubic  inches.     What  is  the  volume  of  a  sphere  whose 
radius  is  10  inches,  if  it  be  known  that  the  volume  varies  as 
the  cube  of  the  radius  ? 


26]  VARIATION.  323 

It  has  been  found  by  experiment  that  the  distance  a  body 
falls  from  rest  varies  as  the  square  of  the  time. 

10.  If  a  body  falls  256  feet  in  4  seconds,  how  far  will  it  fall 
in  10  seconds  ? 

11.  From  what  height  must  a  body  fall  to  reach  the  earth 
after  15  seconds  ? 

It  has  been  found  by  experiment  that  the  velocity  acquired 
by  a  body  falling  from  rest  varies  as  the  time. 

12.  If    the   velocity   of  a  falling  body   is   160   feet   after 
5  seconds,  what  will  be  the  velocity  after  8  seconds  ? 

13.  How  long  must  a  body  have  been  falling  to  have  acquired 
a  velocity  of  256  feet  ? 

14.  The  surface  of  a  cube  whose  edge  is  5  inches  is  150  square 
inches.     What  is  the  surface  of  a  cube  whose  edge  is  9  inches, 
if  it  be  known  that  the  surface  varies  as  the  square  of  its  edge  ? 

15.  It  has  been  found  by  experiment  that  the  weight  of  a 
body  varies  inversely  as  the  square  of  its  distance  from  the 
centre  of  the  earth.     If  a  body  weighs  30  pounds  on  the  surface 
of  the  earth  (approximately  4000  miles  from  the  centre),  what 
would  be  its  weight  at  a  distance  of  24,000  miles  from  the 
surface  of  the  earth  ? 

It  has  been  found  by  experiment  that  the  illumination  of  an 
object  varies  inversely  as  the  square  of  its  distance  from  the 
source  of  light. 

16.  If  the  illumination  of  an  object  at  a  distance  of  10  feet 
from   a   source   of  light   is  2,  what  is  the  illumination  at  a 
distance  of  40  feet  ? 

17.  To  what  distance  must  an  object  which  is  now  10  feet 
from  a  source  of  light  be  removed  in  order  that  it  shall  receive 
only  one-half  as  much  light  ? 

18.  At  what  distance  will  a  light  of  intensity  10  give  the 
same  illumination  as  a  light  of  intensity  8  gives  at  a  distance 
of  50  feet  ? 


CHAPTER   XXI. 

PROGRESSIONS. 

1.  A  Series  is  a  succession  of  numbers,  each  formed  accord- 
ing to  some  definite  law.     The  single  numbers  are  called  the 
Terms  of  the  series. 

E.g.,  in  the  series 

1+3 +  5 +  7 +  9  +  -.  (1) 

each  term  after  the  first  is  formed  by  adding  2  to  the  preced- 
ing term. 

In  the  series  1  +  2  +  4  +  8  H (2) 

each  term  after  the  first  is  formed  by  multiplying  the  preced- 
ing term  by  2. 

2.  The  number  of  terms  in  a  series  may  be  either  limited  or 
unlimited. 

A  Finite  series  is  one  of  a  limited  number  of  terms. 
An  Infinite  series  is  one  of  an  unlimited  number  of  terms. 
In  this  chapter  a  few  simple  'and  yet  very  important  series 
will  be  discussed. 

ARITHMETICAL  PROGRESSION. 

3.  An  Arithmetical  Series,  or,  as  it  is  more  commonly  called, 
an  Arithmetical  Progression  (A.  P.),  is  a  series  in  which  each 
term,  after  the  first,  is  formed  by  adding  a  constant  number  to 
the  preceding  term.     See  Art.  1,  (1). 

4.  Evidently  this  definition  is  equivalent  to  the  statement, 
that  the   difference   between   any   two   consecutive   terms   is 
constant. 

E.g.,  in  the  series 

1+3  +  5  +  7  +  -.. 

we  have  3-1  =  5-3  =  7-5=... 

324 


1-8]  ARITHMETICAL   PROGRESSION.  325 

For  this  reason  the  constant  number  of  the  first  definition  is 
called  the  Common  Difference  of  the  series. 

5.  Let     al  stand  for  the  first  term  of  the  series, 

an  for  th  nth  (any)  term  of  the  series, 

d  for  the  common  difference, 
and  Sn  for  the  sum  of  n  terms  of  the  series. 

The  five  numbers  Oj,  an,  d,  n,  Sn  are  called  the  Elements  of 
the  progression. 

6.  The  common  difference  may  be  either  positive  or  negative. 
If  d  be  positive,  each  term  is  greater  than  the  preceding,  and 

the  series  is  called  a  rising,  or  an  increasing  progression. 

E.g.,  1  +  2+3+4-1-  ..-,  wherein  d  =  1. 

If  d  be  negative,  each  term  is  less  than  the  preceding,  and 
the  series  is  called  a  falling)  or  a  decreasing  progression. 

E.g.,  1  —  1-3-5—  •••,  wherein  d  =  —  2. 

The  nth  Term  of  an  Arithmetical  Progression. 

7.  By  the  definition  of  an  arithmetical  progression, 

a:  =  Oj,  a2  =  al  +  d,  a3  =  a2  +  d  =  ax  +  2  d,  etc. 
The  law  expressed  by  the  formulae  for  these  first  three  terms 
is  evidently  general,  and  since  the  coefficient  of  d  in  each  is 
one  less  than  the  number  of  the  corresponding  term,  we  have 
an  =  a,  +  (n  -  !)</.  (I.) 

That  is,  to  find  the  nth  term  of  an  arithmetical  progression : 
Multiply  the  common  difference  by  n  —  1,  and  add  the  product  to 
the  first  term.  t 

8.  Ex.  1.    Find  the  15th  term  of  the  progression, 

1  +  3  +  5  +  7  +  -. • 

We  have  al  =  1,  d  =  2,  n  =  15; 

therefore  a15  =  1  +  (15  -  1)2  =  1  +  28  =  29. 

This  formula  may  be  used  not  only  to  find  an,  when  aly  d, 
and  n  are  given,  but  also  to  find  any  one  of  the  four  numbers 
involved  when  the  other  three  are  given. 


326  ALGEBRA.  [Cn.  XXI 

Ex.  2.   If  a5  =  8(71  =  5),  and  %  =  1,  we   have  3  =  1  +  4  d  ; 
whence 


The  Sum  of  n  Terms  of  an  Arithmetical  Progression. 

9.  The  successive  terras  in  an  arithmetical  progression,  from 
the  first  to  the  nth  inclusive,  may  be  obtained  either  by  repeated 
additions  of  the  common  difference  beginning  with  the  first 
term,  or  by  repeated  subtractions  of  the  common  difference 
beginning  with  the  nth  term.  We  may  therefore  express  the 
sum  of  n  terms  in  two  equivalent  ways  : 


Whence,  by  addition, 

2Sn  =  (a,  +  O  +  (a,  +  aB)  +  »•  +  (aj  +  ««)  +  K  +  an), 
wherein  there  are  n  binomials,  ax  -f-  an. 

Therefore,  2  Sn  =  n  (a,  +  an),  or  Sn  =  jjj  (a!  +  an).  (II.) 

10.  If  the  value  of  an,  given  in  (I.),  be  substituted  for  a}  in 
(II.),  we  obtain 

Sa  =  ^at  +  (n-l)d].  (III.) 

Formula  (II.)  is  used  when  aa,  art,  and  n  are  given ;  and  (III.) 
when  a1?  d,  and  n  are  given. 

11.  Ex.1.   If  a1  =  l,  a5  =  3,  then /S5  =  f  (1  +  3)  =  10. 
Ex.  2.    If  «!  =  —  4,  d  =  2,  n  =  12, 

then  S12  =  if.  [2  ( -  4)  4-  ll  x  2]  =  84. 

Either  (II.)  or  (III.)  can  be  used  to  determine  any  one  of 
the  five  elements  al5  an,  d,  n,  Sn,  when  the  three  others  involved 
in  the  formula  are  known. 

Ex.  3.    Given  a1  =  —  3,d  =  2,  Sn  =  12,  to  find  n. 
From  (III.),  12  =  |[-6  +  2(n-l)], 
or  n2  —  4  n  =  12 ;  whence  n  =  6  and  —  2. 


8-12]  ARITHMETICAL   PROGRESSION.  327 


The  result  6  gives  the  series  —3  —  1  +  1  +  3  +  5  +  7,  =12. 

Since  the  number  of  terms  must  be  positive,  the  negative 
result,  —  2,  is  not  admissible.  But  its  meaning  may  be 
assumed  to  be  that  two  terms,  beginning  with  the  last  and 
counting  toward  the  first,  are  to  be  taken. 

12.  Formulae  (I.)  and  (II.),  or  (I.)  and  (III.),  may  be  used 
simultaneously  to  determine  any  two  of  the  five  numbers  aly 
an,  d,  Sn,  n  when  the  three  others  are  given. 

Ex.    Given    d  =  —2,   an  =  -  16,  Sn  =  -  60,  to  find  a^  and  n. 
From  (I.),  -  16  =  a,  -  2  (n  -  1),  (1) 

and  from  (II.),  -  60  =  -  (%  -  16).  (2) 

2 

Solving  (1)  and  (2),  we  obtain  n  =  12,  «j  =  6  ;  and  n  =  5, 
a-!  =  -  8. 

The  two  series  are  : 

6  +  4  +  2  +  0-2-4-6-8-10-12-14-16, 
and  -  8  -  10  -  12  -  14  -  16, 

both  of  which  have  d  =  —2,  an  =  -  16,  Sn  =  —  60. 

Notice  that  in  this  example  the  sum  of  the  terms  which  are 
not  common  to  the  two  series  is  0. 

EXERCISES  I. 

Find  the  last  term,  and  the  sum  of  the  terms,  of  each  of  the 
following  arithmetical  progressions  : 

1.  2  +  6H  ----  to  10  terms.          2.   3  +  1  ----  to  13  terms. 

3.  _5_  2+  ..-to  21  terms.       4.   3  +  11  +  •••  to  40  terms. 

5.  4  +  If  ----  to  31  terms.       6.    9  +  11  H  ----  to  n  terms. 

7.  n  +  2n  +  •••  to  16  terms,  to  m  terms. 

8.  a  +  (a  +  &)+•••  to  20  terms,  to  n  terms. 

9.  (m  +  2)  +  (4  m  +  5)  +  -"to  40  terms,  to  n  terms. 

10.    £_1  +  a~  3  +  ...  to  30  terms,  to  n  terms. 


328  ALGEBRA.  [Cii.  XXI 

In  each  of  the  following  arithmetical  progressions  find  the 
values  of  the  two  elements  not  given  : 

11.  «!  =  4,  d  =  5,  n  =  10.  12.  aB  =  16,  d  =  2,  n  =  9. 

13.  0!  =  2t,  /i=5,  am=-1.9.  14.  d  =  -4.8,  n=3,  £n=28.5. 

15.  an=13,  w  =  8,  £ro=100.  16.  an  =  2j,  ?i  =  12,  Sn  =  -7. 

17.  a1==9,  d=-l,  an=6.  18.  ^  =  22$,  an=-19f,  £B=20. 

19.  al=2,  d=5,  £n=245.  20.  an=56,  d=5,  £n= 


Arithmetical  Means. 

13.  The  Arithmetical  Mean  between  two  numbers  is  a  third 
number,  in  value  between  the  two,  which  forms  with  them  an 
arithmetical  progression. 

E.g.,  2  is  an  arithmetical  mean  between  1  and  3. 
Let  A  stand  for  the  arithmetical  mean  between  a  and  bj 
then,  by  the  definition  of  an  arithmetical  progression, 
A  —  a  =  b  —  A, 

whence  A  —  ^~~  — 

That  is,  the  arithmetical  mean  between  two  numbers  is  half 
their  sum. 

14.  Arithmetical  Means  between  two  numbers  are  numbers, 
in  value  between  the  two,  which  form  with  them  an  arithmeti- 
cal progression. 

E.g.,  2,  3,  and  4  are  three  arithmetical  means  between  1  and  5. 
Ex.   Insert  four  arithmetical  means  between  —  2-  and  9. 
We  have  n  =  6,  al  =  —  2,  «6  =  9. 

From  (I.),         9  =  -  2  +  5  d,  whence  d  =  -U- 
The  required  means  are  £,  JJ2-,  -2/-,  ^-. 

EXERCISES  II. 

Insert  an  arithmetical  mean  between 

1.   45  and  31.  2.   17£  and  14f  3.   2  a  and  -  2  b. 

and  «±^.  5.   ^±1  and  -±1- 


a  +  b          a  —  b  x  —  l 


12-16]  ARITHMETICAL   PROGRESSION.  329 

6.  Insert  six  arithmetical  means  between  7  and  35. 

7.  Insert  twelve  arithmetical  means  between  37  and  —  28. 

8.  Insert  nine  arithmetical  means  between  -J-  and  12. 

9.  Insert  twenty  arithmetical  means  between  —  16  and  26. 
10.  Insert  six  arithmetical  means  between  a+b  and  8  a  — 13  b. 

Problems. 

15.  Pr.     Find  the  sum  of  all  the  numbers  of  three  digits 
which  are  multiples  of  7. 

The  numbers  of  three  digits  which  are  multiples  of  7  are 

7x15,  7x16,  7  x  17,  -.,7x142. 
Their  sum  is          7(15  +  16  +  -  •  •  +  142). 

The  series  within  the  parentheses  is  an  arithmetical  progres- 
sion, in  which  ax  =  15,  d  =  1,  n  =  128,  and  am  =  142. 

Therefore  S128  =  10048. 

The  required  sum  is  therefore  7  x  10048,  =  70336. 

16.  In  many  examples  the  elements  necessary  for  determin- 
ing the  required  element  or  elements  directly  from  (L)-(III.) 
are  not  given,  but  in  their  place  equivalent  data.       , 

Ex.  1.  The  sixth  term  of  an  A.  P.  is  17,  and  the  eleventh 
term  is  32.  Find  the  first  term  and  the  common  difference. 

We  have  a6  =  17,  an  =  32. 

From  (I.),     17  =  Oj  +  5  d,  and  32  =  ax  +  10  d. 

Solving  these  equations,  Oj  =  2.  d  =  3. 

Or  we  could  have  regarded  17  as  the  first  term  and  32  as 
the  last  term  of  a  progression  of  six  terms.  Then,  by  (I.), 
32  =  17  -}-  5  d,  whence  d  =  3. 

By  (I.)  again,  17  =  a:  +  5  x  3 ;  whence  (^  =  2,  as  above. 

EXERCISES  III. 

1.  Find  the  sixth  term,  and  the  sum  of  eleven  terms,  of  an 
A.  P.  whose  eighth  term  is  11  and  whose  fourth  term  is  —  1. 


330  ALGEBRA.  [Cn,  XXI 

2.  The  sixteenth  term  of  an  A.  P.  is  —  5,  and  the  forty-first 
term  is  45.     What  is  the  first  term,  and  the  sum  of  twenty 
terms  ? 

3.  Find  the  sum  of  all  the  even  numbers  from  2  to  50 
inclusive. 

4.  Find  the  sum  of  thirty  consecutive   odd  numbers,   of 
which  the  last  is  127. 

5.  The  sum  of  the  eighth  and  fourth  terms  of  an  A.  P.  of 
twenty  terms  is  24,  and  the  sum  of  the  fifteenth  and  nineteenth 
terms  is  68.     What  are  the  elements  of  the  progression  ? 

6.  The  sum  of  the  second  and  twentieth  terms  of  an  A.  P. 
is  10,  and  their  product  is  23  JJ.     What  is  the  sum  of  sixteen 
terms  ? 

7.  The  sixth  term  of  an  A.  P.  is  30,  and  the  sum  of  the 
first  thirteen  terms  is  455.     What  is  the  sum  of  the  first  thirty 
terms  ? 

8.  What  value  of  x  will  make  the  arithmetical  mean  be- 
tween x*  and  X*  equal  to  6  ? 

9.  Find  the  sum  of  all  even  numbers  of  two  digits. 

10.  How  many  consecutive  odd  numbers  beginning  with  7 
must  be  taken  to  give  a  sum  775  ? 

11.  Insert  between  0  and  6  a  number  of  arithmetical  means 
so  that  the  sum  of  the  terms  of  the  resulting  A.  P.  shall  be  39. 

12.  Find  the  number  of  arithmetical  means  between  1  and 
19,  if  the  second  mean  is  to  the  last  mean  as  1  to  7. 

13.  The  sum  of  the  terms  of  an  A.  P.  of  six  terms  is  66,  and 
the  sum  of  the  squares  of  the  terms  is  1006.     What  are  the 
elements  of  the  progression  ? 

14.  The  sum  of  the  terms  of  an  A.  P.  of  twelve  terms  is  354, 
and  the  sum  of  the  even  terms  is  to  the  sum  of  the  odd  terms 
as  32  to  27.     What  is  the  common  difference  ? 

15.  How  many  positive  integers  of   three  digits  are  there 
which  are  divisible  by  9  ?     Find  their  sum. 


16-19]  GEOMETRICAL  PROGRESSION.  331 

16.  Show  that  the  sum  of  2  n  -f-  1  consecutive  integers  is 
divisible  by  2n  •+- 1. 

17.  Prove  that  if  the  same  number  be  added  to  each  term  of 
an  A.  P.,  the  resulting  series  will  be  an  A.  P. 

18.  Prove  that  if  each  term  of  an  A.  P.  be  multiplied  by  the 
same  number,  the  resulting  series  will  be  an  A.  P. 

19.  Prove  that  if  in  the  equation  y  =  ax  +  6,  we  substitute 
c,  c  +  d,  c  +  '2d,  •  ••,  in  turn  for  x,  the  resulting  values  of  y  will 
form  an  A.  P. 

20.  A  laborer  agreed  to  dig  a  well  on  the  following  con- 
ditions :  for  the  first  yard  he  was  to  receive  $  2,  for  the  second 
$  2.50,  for  the  third  $  3,  and  so  on.     If  he  received  $  42.50  for 
his  work,  how  deep  was  the  well  ? 

21.  On  a  certain  day  the  temperature  rose  1°  hourly  from  5 
to  11  A.M.,  and  the  average  temperature  for  that  period  was  8°. 
What  was  the  temperature  at  8  A.M.  ? 

22.  Twenty-five  trees  are  planted  in  a  straight  line  at  inter- 
vals of  5  feet.     To  water  them,  the  gardener  must  bring  water 
for  each  tree  separately  from  a  well  which  is  10  feet  from  the 
first  tree  and  in  line  with  the  trees.     How  far  has  the  gardener 
walked  when  he  has  watered  all  the  trees  ? 

GEOMETRICAL  PROGRESSION. 

17.  A  Geometrical  Series,  or,  as  it  is  more  commonly  called, 
a  Geometrical  Progression  (G.  P.),  is  a  series  in  which  each  term 
after  the  first  is  formed  by  multiplying  the  preceding  term  by 
a  constant  number.     See  Art.  1,  (2). 

18.  Evidently  this  definition  is  equivalent  to  the  statement 
that  the  ratio  of  any  term  to  the  preceding  is  constant. 

For  this  reason  the  constant  multiplier  of  the  first  definition 
is  called  the  Ratio  of  the  progression. 

19.  Let  a^  stand  for  the  first  term  of  the  series, 

an  for  the  nth  (any)  term, 
r  for  the  ratio, 
and  Sn  for  the  sum  of  n  terms. 


332  ALGEBRA.  [Cn.  XXI 

20.  The  ratio  may  be  either  larger  or  smaller  than  1;  in 
the  former  case  the  progression  is  called  a  rising  or  ascending 
progression  ;  in  the  latter  a  falling  or  descending  progression. 

E.g.,  1  +  I  +  .|  +  2JL  +  ...?  in  which  r  =  f, 

and  £  —  1  +  2  —  4  +  8  —  ,  in  which  r  =  -  2, 

are  ascending  progressions  ;  while 

1  +  i  +  i  +  1  +  *••>  in  which  r  =  i> 
and  1  —  -§-  -J-  f  —  -£r  +  •  •  •>  in  which  r  =  —  f  , 

are  descending  progressions. 

The  flth  Term  of  a  Geometrical  Progression. 

21.  By  the  definition  of  a  geometrical  progression 

al  =  a1?  a2  =  c^r,  a3  =  a>2r  =  a^,  a4  =  a3r  =  c^r3,  etc. 
The  law  expressed  by  the  relations  for  these  first  four  terms 
is  evidently  general,  and  since  the  exponent  of  r  in  each  is  one 
less  than  the  number  of  the  corresponding  term,  we  have 

*„  =  a^~1'  (L) 

That  is,  to  find  the  nth  term  of  a  geometrical  progression  : 
Raise  the  ratio  to  a  power  one  less.  than  the  number  of  the  term, 
and  multiply  the  result  by  the  first  term. 

Ex.  1.    If  al  =  i,  r  =  3,  n  =  5,  then  a5  =  \  -  34  =  -8J-. 

This  relation  may  also  be  used  to  find  not  only  an,  when  a1? 
r,  and  n  are  given,  but  also  to  find  the  value  of  any  one  of  the 
four  numbers  when  the  other  three  are  given. 

Ex.  2.    If  %  =  4,  a6  =  i,  n  =  6,  then  -J-  =  4  r5,  whence  r  =  £. 

The  Sum  of  a  Geometrical  Progression. 

22.  We  have  Sn=a1  +  a1r+alr2-\  -----  ha1rn-2+a1rn-1,  (1) 
and                     rJSn  =        axr  -f  a^2  H  -----  f-  a^"-2  -f  a^"-1  -f-  a^n.  (2) 

Consequently,  subtracting  (2)  from  (1), 
#n(l  -  r)  =  «!  -  a^", 


whence  Sn  =         -       =          -.  (IL) 

1  —  r  r  —  1 


20-24]  GEOMETRICAL   PROGRESSION.  333 

Substituting  an  for  a^M~l  in  (II.),  we  have 

Sn   =  «1  ~    «!»/*  =  g^-fll.  (HI.) 

1  —  /•  /*  —  1 

The  first  forms  of  (II.)  and  (III.)  are  to  be  used  when  r  <  1, 
the  second  when  ?•  >  1. 

23.   Ex.  1.    Given  «x  =  3,  r  =  2,  w  =  6,  to  find  £6. 
From  (II.),  #6  =  3(|j"l1)  =  189. 

Formulae  (II.)  and  (III.)  may  be  used  not  only  to  find  Sn 
when  Oi,  r,  and  n,  or  al5  aw,  and  r  are  given,  but  also  to  find  the 
value  of  any  one  of  the  four  numbers  when  the  other  three 
are  given. 

Ex.  2.    Given  Sn  =  -  63},  «a  =  -  },  aB  =  —  32,  to  find  r. 
By  (III.),   -  63}  =  ~     +  32r,  whence  r  =  2. 


24.  Formulae  (I.)  and  (II.),  or  (I.)  and  (III.),  may  be  used 
simultaneously  to  determine  any  two  of  the  five  elements, 
«!,  «•„,  r,  £n,  n,  when  the  three  other  elements  are  given. 

Ex.    Given  r  =  2,  an  =  16,  Sn  =  31},  to  find  ^  and  n. 

From  (III.),  31}  =  16*2~ai,  whence  %  =  }. 
^  —  1 

From  (I.),        16  =}  •  2n~1,  whence  n  =  6. 
EXERCISES  IV. 

Find  the  last  term  and  the  sum  of  the  terms  of  each  of  the 
following  geometrical  progressions  : 

1.    3  +  6  +  •••  to  six  terms.  2.    2  —  4  -f-  -••  to  ten  terms. 

3.    32  —  1.6  +  •••  to  seven  terms.       4.    If  -f  2|-f  •••  to  six  terms. 

2        1 
5.    2  —  22  +  ••  to  eleven  terms.      6.    -^  +^  +  •••  to  n  terms. 

7.    1  +  (1  +  »)  +  •••  to  four  terms,  to  n  terms. 


334  ALGEBRA.  [Cn.  XXI 

In  each  of  the  following  geometrical  progressions  find  the 
values  of  the  elements  not  given  : 

8.  %  =  1,  r  =  4,  n  =  5.  9.  an  =  10,  r  =  2,  n  =  4. 

10.  an=96,  w=4,  Sn=127.5.  11.  r=10,  n=7,  #B=3,333,333. 

12.  Oi  =  74|,  n  =  6,  an  =  21.  13.  a:  =  7,  r  =  10,  an  =  700. 

14.  «!  =  !,  aB=512,  #B  =  1023.  15.  an  =  3125,  r  =  5,  £n  =  3905. 

16.  a!  =4,  r=3,  £n=118,096.  17.  ax  =  100,  w  =  3,  £„  =  700. 

25.  The  Sum  of  an  Infinite  Geometrical  Progression.  —  If  the 
number  of  terms  in  a  geometrical  progression  is  unlimited, 
the  exact  value  of  the  sum  of  the  series  cannot  be  obtained. 
Thus,  in  the  series 

1+i+i+iH  ----  without  end, 

the  sum  continually  increases  as  more  and  more  terms  are 
included  in  it. 


=2  -ay-1. 

And  S,  =  1,  S2  =  11   S3  =  1},  ^4  =  1J,  •»• 

^1000=  2  —(I)999;  and  so  on. 

We  thus  see  that,  although  the  sum  of  this  series  grows 
larger  and  larger,  it  does  not  increase  without  limit,  but  ap- 
proaches the  value  2  more  and  more  nearly  as  more  and  more 
terms  are  included  in  the  sum.  Evidently  the  sum  can  be 
made  to  differ  from  2  by  as  little  as  we  please,  by  taking  a 
sufficient  number  of  terms. 

We  therefore  call  2  the  limit  of  the  sum  of  the  series,  or  more 
briefly,  the  sum  of  the  series.  The  exact  sum  2,  however,  can 
never  be  obtained. 

26.   In  general,  when  r  <  1,  the  term  0$*  in  the  formula 


24-26]  GEOMETRICAL  PROGRESSION.  335 

decreases  as  n  increases.  It  can  be  proved,  as  in  the  particular 
example,  that  this  term  can  be  made  as  small  as  we  please,  by 
taking  n  sufficient!}7  great. 

Therefore,  when  r  <  1,  we  take 


as  the  sum  of  the  infinite  geometrical  progression. 

This  theory  can  be  applied  to  find  the  value  of  a  repeating 
(recurring)  decimal. 

Ex.   Verify  that  .6  =  |. 

We  have        .666  ...  =  T%  +  T^  +  ^  +  -, 

a  geometrical  progression  whose  first  term  is  •£$  and  whose 
ratio  is  -^.     Consequently 

Q  _       TU"        _  6  _  2 
°  -  1_     1    -  TT  -  3  ' 

*-       10 
EXERCISES  V. 

Find  the  sum  of  the  following  infinite  geometrical  progres- 
sions : 
1.   6  +  4  +  ..-.       2.   60  +  15  +  ..-.          3.   10-6  +  .... 

4-   i  +  i  +  --       5-  l-i-H-  ••  6.  5-1  +  .... 

7.    f-f  +  ....        8.    Vf_f.V2+...t 

10.  l  +  £  +  &i2  +  .«-,  when 

11.  l+-  +  iH  ----  ,  when 

x     x2 

Find  the  value  of  each  of  the  following  repeating  decimals  : 
12.   .44....  13.    .99..-.  14.   .2727-... 

15.   .015015-...  16.   .199....  17.   1.0909-  ... 

18.    .122323-...  19.   .2014T5475-... 

Verify  each  of  the  following  identities  : 
20. 


336  ALGEBRA.  [Cii.  XXI 

Geometrical  Means. 

27.  A  Geometrical  Mean  between  two  numbers  is  a  number, 
in  value  between  the  two,  which  forms  with  them  a  geometrical 
progression. 

E.g.,  +  2,  or  —  2,  is  a  geometrical  mean  between  1  and  4. 

Let  G  be  the  geometrical  mean  between  a  and  b. 

Then  by  definition  of  a  geometrical  progression, 

•^=  -|;  whence  G  =  ±  V(«6). 

TJiat  is,  the  geometrical  mean  between  two  numbers  is  the 
square  root  of  their  product. 

Ex.   Find  the  geometrical  mean  between  1  and  -J.     We  have 


28.  Geometrical  Means  between  two  numbers  are  numbers,  in 
value  between  the  two,  which  form  with  them  a  geometrical 
progression.  E.g.,  4  and  16  are  two  geometrical  means  between 
1  and  64;  and  2,  4,  8,  16,  32  are  five  geometrical  means 
between  1  and  64. 

Ex.   Insert  five  geometrical  means  between  1  and  729. 
We  have  al  =  1,  n  =  7,  a,  =  729. 

Therefore  729  =  r6,  or  r  =  ±  3. 

The  required  means  are  : 

±  3,  9,  ±  27,  81,  ±  243. 

EXERCISES   VI. 

Insert  a  geometrical  mean  between 
1.    2  and  8.  2.    12  and  3.  3.   -J-  and  yfy. 

4.    ^/a  and  V  (2  a).       5.    75  m3  and  3  mn*.      6.    —  and  —  . 

q          p 

7.    (a-6)2and(a+6)2.        8.    (a2  +1)  (a2-!)-1  and  \  (a4-!). 
9.   Insert  five  geometrical  means  between  2  and  1458. 
10.   Insert  seven  geometrical  means  between  2  and  512. 


27-30]  GEOMETRICAL   PROGRESSION.  337 

11.  Insert  six  geometrical  means  between  3  and  —  384. 

12.  Insert  six  geometrical  means  between  5  and  —  640. 

13.  Insert  nine  geometrical  means  between  1  and 


Problems. 

29.  Pr.  A  farmer  agrees  to  sell  12  sheep  on  the  following 
terms  :  he  is  to  receive  2  cents  for  the  first  sheep,  4  cents  for 
the  second,  8  cents  for  the  third,  and  so  on.  How  much  does 
he  receive  for  the  twelfth  sheep,  and  how  much  for  the  12 
sheep,  and  what  is  the  average  price  ? 

We  have  ax  =  2,n  =  12,  r  =  2. 

Then  a12  ==  2  x  211  =  212  =  4096. 


And  #12  =          ~      =  2  x  4095  =  8190. 

That  is,  he  receives  4096  cents,  or  $  40.96,  for  the  twelfth 
sheep,  and  8190  cents,  or  $  81.90,  for  the  12  sheep. 

The  average  price  is  ^&,  =  $  6.82  J. 
1^ 

30.  In  many  examples  the  elements  necessary  for  determin- 
ing the  element  or  elements  directly  from  (I.)-(IIL)  are  not 
given,  but  in  their  place  equivalent  data. 

Ex.  The  fifth  term  of  a  G.  P.  is  48,  and  the  eighth  term  is 
384.  Find  the  first  term  and  the  ratio. 

From  (I.),  48  =  a,r4,  and  384  =  a,r7  ; 

whence  r8=  8,  or  r  =  2.     Therefore  ax  =  3. 

Or,  we  could  have  regarded  48  as  the  first  term  and  384  as 
the  last  term  of  a  progression  of  four  terms.  Then  by  (I.), 
384  =  48  r3,  whence  r  =  2  as  before. 

EXERCISES   VII. 

1.  .The  first  term  of  a  G.  P.  of  six  terms  is  768,  and  the  last 
term  is  one-sixteenth  of  the  fourth  term.  What  is  the  sum  of 
the  six  terms  of  the  progression? 


338  ALGEBRA.  [Cn.  XXI 

2.  The  first  term  of  a  G.  P.  of  ten  terms  is  3,  and  the  sum 
of  the  first  three  terms  is  one-eighth  of  the  sum  of  the  next 
three  terms.     Find  the  elements  of  the  progression. 

3.  The  twelfth  term  of  a  G.  P.  is  1536,  and  the  fourth  term 
is  6.    What  is  the  ratio,  and  the  sum  of  the  first  eleven  terms  ? 

4.  In  a  G.  P.  of  eight  terms,  the  sum  of  the  first  seven 
terms  is  4441,  and  is  to  the  sum  of  the  last  seven  terms  as 
1  to  2.     Find  the  elements  of  the  progression. 

5.  The  sum  of  the  first  four  terms  of  a  G.  P.  is  15,  and  the 
sum  of  the  terms  from  the  second  to  the  fifth  inclusive  is  30. 
What  is  the  first  term,  and  the  ratio  ? 

6.  Find  the  elements  of  a  G.  P.  of   six  terms  whose  first 
term  is  1,  and  the  sum  of  whose  first  six  terms  is  28  times  the 
sum  of  the  first  three  terms. 

7.  The  sum  of  the  first  three  terms  of  a  G.  P.  is  21,  and  the 
sum  of  their  squares  is  189.     What  is  the  first  term  ? 

8.  The  product  of  the  first  three  terms  of  a  G.  P.  is  216, 
and  the  sum  of  their  cubes  is  1971.     What  is  the  first  term, 
and  the  ratio  ? 

9.  If  the  numbers  1,  1,  3,  9  be  added  to  the  first  four  terms 
of  an  A.  P.,  respectively,  the  resulting  terms  will  form  a  G.  P. 
What  is  the  first  term,  and  the  common  difference  of  the  A.  P.  ? 

10.  A  G.  P.  and  an  A.  P.  have  a  common  first  term  3,  the 
difference   between   their   second  terms  is  6,  and  their  third 
terms  are  equal.     What  is  the  ratio  of  the  G.  P.,  and  the  com- 
mon difference  of  the  A.  P.  ? 

11.  Show  that,  if  all  the  terms  of  a  G.  P.  be  multiplied  by 
the  same  number,  the  resulting  series  will  form  a  G.  P. 

12.  Show  that  the  series  whose  terms  are  the  reciprocals  of 
the  terms  of  a  G.  P.  is  a  G.  P. 

13.  Show  that  the  product  of  the  first  and  last  terms  of  a 
O.  P.  is  equal  to  the  product  of  any  two  terms  which  are 
.equally  distant  from  the  first  and  last  terms  respectively. 


30-34]  HARMONICAL   PROGRESSION.  339 

14.  A  merchant's   investment   yields  him  each  year   after 
the  first,  three  times  as  much  as  the  preceding  year.     If  his 
investment  paid  him  $  9720  in  four  years,  how  much  did  he 
realize  the  first  year  and  the  fourth  year  ? 

15,  Given  a  square  whose  side  is  2  a.     The  middle  points 
of   its  adjacent  sides  are  joined  by  lines  forming   a  second 
square   inscribed  in   the  first.     In  the  same  manner  a  third 
square  is  inscribed  in  the  second,  a  fourth  in  the  third,  and  so 
on   indefinitely.     Find  the  sum  of  the  perimeters  of   all  the 
squares. 

HARMONICAL   PROGRESSION. 

31.  A  Harmonical  Progression  (H.  P.)  is  a  series  the  recipro- 
cals of  whose  terms  form  an  arithmetical  progression. 

E-9-,  l  +  i  +  i  +  i+-- 

is  a  harmonical  progression,  since 

1  +  2  +  3  +  4+.. • 

is  an  arithmetical  progression. 

Consequently  to  every  harmonical  progression  there  corre- 
sponds an  arithmetical  progression,  and  vice  versa. 

32.  Any  term  of  a  harmonical  progression  is  obtained  by 
finding  the  same  term  of  the  corresponding  arithmetical  pro- 
gression and  taking  its  reciprocal. 

Ex.  Find  the  eleventh  term  of  the  harmonical  progression 
4,2,|,.... 

The  corresponding  arithmetical  progression  is 

i  *>  i,  •-, 

and  its  eleventh  term  is  J^-. 

Therefore  the  eleventh  term  of  the  given  progression  is  T4T. 

•     33.  No  formula  has  been  derived  for  the  sum  of  n  terms  of 
a  harmonical  progression. 

34.  A  Harmonical  Mean  between  two  numbers  is  a  number,, 
in  value  between  the  two,  which  forms  with  them  a  harmonical 
progression. 


340  ALGEBRA.  [Cn.  XXL 

E.g.,  f  is  a  harmonical  mean  between  ^  and  —  f . 

Let  H  stand  for  the  harmonical  mean  between  a  and  ft,  then 

—  is  an  arithmetical  mean  between  -  and  —     Consequently 

M 

1      a     b         „      2ab 

-,  or  H= -. 


-Ll  £ 

Ex.    Insert  a  harmonical  mean  between  2  and  5. 

We  have  g=2*2x5  =  20. 

2  +  5          7 

35.  Harmonical  Means  between  two  numbers  are  numbers,  in 
value  between  the  two,  which  form  with  them  a  harmonical 
progression. 

E.g.,  f,  1,  J,  |,  i  are  five  harmonical  means  between  3  and  f . 
Ex.    Insert  four  harmonical  means  between  1  and  10. 
We  have  first  to  insert  four  arithmetical  means  between  1 
and  Y1^,  and  obtain 

TTP  ft?  If?  if • 
The  required  harmonical  means  are  therefore 

50       50       50       50 

if*  fib  ft>  ft- 

Problems. 

36.  Pr.  1.    The  geometrical  mean  between  two  numbers  is  -^, 
and  the  harmonical  mean  is  fi     What  are  the  numbers? 

Let  x  and  y  represent  the  two  numbers. 

Then  V<X?/)  =  •£•>  or  ®y  =  i  >  (1) 


and  =   ,  or  5  xy  =  x  +  y.  (2) 

x  -\-  y      o 

Solving  (1)  and  (2),  we  obtain  x  =  1  ,  y  =  J,  and  #  =  ^,  y  =  1. 

EXERCISES   VIII. 

Find  the  last  term  of  each  of  the  following  harmonical  pro- 
gressions : 

1.    1+    +  1+..  .  to  8  terms.       2.      ++      +  ...  to  15  terms. 


3.    2-2-f  ----  to  11  terms.     4.    8-f-T*T  ----  to  16  terms. 


34-36]  HARMONICAL   PROGRESSION.  341 

5.   i  +  —  -f—  H  ----  to  25  terms. 
a     2  a     3  a 


Find  the  harmonical  mean  between 
7.    2  and  4.  8.    -  3  and  4.  9.   }  and  1. 

10.   _J_  and  --  LJ  11.   «H*  and 


a;  —  1  x  +  1  a-j-6          a—  6 

12.  Insert  5  harmonical  means  between  5  arid  i. 

13.  Insert  10  harmonical  means  between  3  and  1. 

14.  Insert  4  harmonical  means  between  —  7  and  ^. 

15.  If  a  be  the  harmonical  mean  between  b  and  c,  prove  that 

a  —  6_g 
6  —  c      c 

16.  The  arithmetical  mean  between  two  numbers  is  6,  and 
the  harmonical  mean  is  -3^.     What  are  the  numbers  ? 

17.  If  one  number  exceeds  another  by  two,  and  if  the  arith- 
metical mean  exceeds  the  harmonical  mean  by  ^,  what  are  the 
numbers  ? 

18.  The  seventh  term  of  a  harmonical  progression  is  y1^-,  and 
the  twelfth  term  is  ^     What  is  the  twentieth  term  ? 

19.  The  tenth  term  of  a  harmonical  progression  is  J,  and  the 
twentieth  term  is      .     What  is  the  first  term  ? 


CHAPTER   XXII. 

THE    BINOMIAL   THEOREM    FOR   POSITIVE    INTEGRAL 
EXPONENTS. 

1.  The  expansions  of  the  powers  of  a  binomial,  from  .the 
third  to  the  fourth  inclusive,  were  given  in  Ch.  XIII.,  Arts. 
7-8,  and  the  laws  governing  the  expansion  of  these  powers 
were  stated. 

As  yet,  however,  we  cannot  infer  that  these  laws  hold  for 
the  fifth  power  without  multiplying  the  expansion  of  the 
fourth  power  by  a  +  b  ;  nor  for  the  sixth  power  without  next 
multiplying  the  expansion  of  the  fifth  power  by  a  +  b;  and 
so  on. 

If,  however,  we  prove  that,  provided  the  laws  hold  for  any 
particular  power,  they  hold  for  the  next  higher  power,  we  can 
infer,  without  further  proof,  that  because  the  laws  hold  for  the 
fourth  power,  they  hold  also  for  the  fifth;  then  that  because 
they  hold  for  the  fifth,  they  hold  also  for  the  sixth,  and  so  on 
to  any  higher  power. 

2.  If  the  laws  (i.)-(vi.)  hold  for  the  rth  power,  we  have 


1  •  2i  •  3 

Notice  that  only  the  first  four  terms  of  the  expansion  are 
written.  But  it  is  often  necessary  to  write  any  term  (the  fcth, 
say)  without  having  written  all  the  preceding  terms. 

To  derive  this  term,  observe  that  the  following  laws  hold  for 
each  term  of  the  expansion  : 

(i.)  The  exponent  of  b  is  one  less  than  the  number  of  the  term 
(counting  from,  the  left). 

Thus  in  the  first  term  we  have  bl~l  =  6°  =  1  ;  in  the  second, 
62-1  =  b  ;  in  the  tenth,  bl°~l  =  b9  ;  and  in  the  &th  term,  ft*-1. 

342 


1-3]  BINOMIAL   THEOREM.  343 

(ii.)  The  exponent  of  a  is  equal  to  the  binomial  exponent  less 
the  exponent  of  b. 

Thus,  in  the  first  term  we  have  ar~°  =  ar ,  in  the  second, 
ar~l ;  in  the  tenth,  ar~9 ;  and  in  the  &th  term,  ar~(k~l),  =  ar~k+l. 

(iii.)  The  number  of  factors  (beginning  with  1  and  increasing 
by  1)  in  the  denominator  of  each  coefficient,  and  the  number  of 
factors  (beginning  with  r  and  decreasing  by  1)  in  the  numerator 
of  each  coefficient,  is  equal  to  the  exponent  ofb  in  that  term. 

Thus,  in  the  coefficient  of  the  second  term  the  denominator 
is  1  and  the  numerator  is  r;  in  that  of  the  third  term  the 
denominator  is  1  •  2  and  the  numerator  is  r(r  —  1);  in  the 
tenth  term  the  denominator  is  1  •  2  •••  9  and  the  numerator  is 
r(r  —  1)  ...  (r  —  8);  and  in  the  &th  term  the  denominator  is 
1  •  2  •  3  •••  (k  —  1),  and  the  numerator  is 

Therefore  the  kth  term  in  the  expansion  of  (a  +  b)r  is 

r(r  _  1)  (r  _  2)  ...  (r  -  k  +  2) ^-fc+i^-i. 

In  like  manner,  any  other  term  can  be  written. 
Thus,  the  (k  —  l)th  term  is 


-a'-k+2bk-2. 


1.2-3  —  (fc-2) 

3.  We  can  now  prove  that,  if  the  laws  (i.)-(vi.)  hold  for 
(a  +  b)r,  they  also  hold  for  (a  +  b)r+l  ;  that  is,  if  they  hold  for 
any  power  they  hold  for  the  next  higher  power.  Assuming, 
then,  that  the  laws  hold  for  (a  -f  6)r,  we  have 


-i   ,   . 
1.  2  V3  ..«•(*  -2)  (&-,!) 

The  first  three  terms  of  the  expansion  are  written,  then  all 
terms  are  omitted,  except  the  (k  —  l)th  and  the  Jfcth. 


344  ALGEBRA.  [Cn.  XXII 

Multiplying  the  expansion  of  (a  -f-  b)r  by  (a  •+•  6),  we  obtain  : 
(a  -f-  b)r+l  =  ar+l  +  rarb  +  r^~1)  ar~lb2  -f  ••• 


f-r(r-l)...(r-fc  +  2)     r(r-l)...(r-fe+3)-| 
1.2...(A;-1)  1.2...(fc-2) 


But 

and 


-  fe  +  3) 


1.2-..(A;-1)  1-2-  »(fc  -2) 

=  r(r  -  1)  -.  (r  -  ft  +  2)  +  r(r  -  1)  —  (r  -  k  +  3)  (ft  -  1) 

1.2...(/b-l) 
r(r  -  1)  ...  (r  -  fe  +  3)  (r  -  fe  +  2  +  ft  -  1) 


Therefore, 
<a  +  b)r+l  =  ar+l  +  (r 

, 


arb 


-f 


!  , 


The  laws  (i.)-(vi.)  hold  for  the  above  expansion  of  (a  -h  b)r+1. 
We  therefore  conclude  that  if  the  expansion  holds  for  (a  +  6)r, 
it  also  holds  for  (a  +  6)r+1. 

Consequently,  since  the  expansion  holds  for  the  fourth  power, 
it  holds  for  the  fifth,  and  so  on  to  any  positive  integral  power. 

The  method  of  proof  employed  in  this  article  is  called  Proof 
by  Mathematical  Induction. 


3-7]  BINOMIAL   THEOREM.  345 

4.   We  may  now  write  the  expansion  of  (a  -f  6)n,  wherein  n 
is  any  positive  integer  : 

(a  +  b)"  =  a"  +  na"-lb  +  -(f^  a"-262  +  .... 

1  •  6 

In  particular,  if  a  =  1,  and  b  =  x, 


. 
1  •  A 

5.   The  expansion  of  (a  —  b)n  can  be  at  once  written  from 
that  of  (a  +  &)n. 

We  have  (a  -  by  =  [a  +  (- 


=  an  4-  nan~l  (—  6)  4- 


=  a  — 


Observe  that  the  signs  of  the  terms  alternate,  +  and  —  , 
beginning  with  the  first,  or  that  the  terms  containing  even 
powers  of  b  are  positive,  and  those  containing  odd  powers  of  b 
are  negative. 

6.  When  n  is  a  positive  integer,  the  number  of  terms  in  the 
expansion  is  limited. 

.,  (a  -f  b)5  =  a«  +  5  a 


5.4.3.2  5.  4.3.2. 

r  r 


5.4.3.2.1.0     1&6 
1.2.3.4.5.6 


The  coefficients  of  the  seventh  and  all  succeeding  terms  con- 
tain 0  as  a  factor.  Therefore  these  terms  drop  out,  and  the 
expansion  ends  with  the  sixth  term.  In  general,  the  expan- 
sion of  (a  +  b)n  ends  with  the  n  -f  1th  term.  For,  the  coeffi- 
cients of  the  n  -f-  2th  and  all  succeeding  terms  contain  n  —  n, 
or  0,  as  a  factor. 

7.  The  expansion  of  (a  4-  b)n  may  also  be  written  in  descend- 
ing powers  of  b. 


346  ALGEBRA.  [Cn.  XXII 

Thus,     (b  +  a)n  =  bn  +  nbn~la  +  n(jl  ~f^bn~2a2  4-—, 

wherein  bn  is  the  last  term  of  the  expansion  given  in  Art.  4, 
n  the  coefficient  of  the  next  to  the  last  term,  and  so  on. 
We  therefore  conclude  : 

In  the  expansion  of  (a  +  &)",  wherein  n  is  a  positive  integer, 
the  coefficients  of  terms  equally  distant  from  the  beginning  and 
end  of  the  expansion  are  equal. 

8.   In  Exs.  1-2  which  follow,  the  coefficients  are  computed 
by  the  principle  given  in  Ch.  XIII.,  Art.  7  (v.). 
Ex.  l.   Expand  (1-2  x2)5. 

We  have  (1  -2x2)5  =  I5  -  5  -  14  -  (2sf)  +  10  •  I3  -  (2  x2)2 
-  10  -  12  .  (2  x2)3  +  5  •  1  •  (2  z2)4  -  (2  x2)5 

=  1  -  10  x2  +  40  x4  -  SOx*  +  SOa8  -  32  x10. 

In  expanding  a  binomial,  the  coefficients  of  the  terms  after 
the  middle  term  may  be  at  once  written  by  the  principle  of 
the  preceding  article.  This  remark  applies  to  the  expansion 
before  it  is  reduced,  as  in  Ex.  1. 

Ex.  2.   Find  the  first  five  terms  of  (a~*  +  ft-2)11. 

We  have 

(a-i  +  6-2)n  =(a-i)ii  +  11  (a-*)10  (2  ft-2)  +  55  (cf^)9(2  b~2)2 


=  a"^  +  22  a~5b~2  +  220  a"  V4  +  1320  a~46~ 


9.   Ex.   Find  the  seventh  term  in  (2  x  —  3y)11. 

In  the  seventh  term  the  exponent  of  —3y(=b)  is  6;  the 
exponent  of  2aj(=a)  is  11  —  6,  =5.  The  denominator  of 
the  coefficient  contains  six  factors  beginning  with  1,  and  the 
numerator  contains  six  factors  beginning  with  11.  Therefore 
the  seventh  term  is 

11.10.9.8.7.6 


1.2.3.4.5. 


(2  x)5(  -  3  ?/)«,  =  10777536  ofy8 . 


7-9]  BINOMIAL   THEOREM.  347 

If  the  second  term  of  the  binomial  is  negative,  it  is  better, 
in  finding  a  particular  term,  to  write  the  binomial  in  the  form 
[a  +  (—  &)],  as  in  the  above  example. 

EXERCISES. 

Write  the  expansion  of  each  of  the  following  powers : 
1.    (a  +  &)6.  2.    (x-y)7.  3.    (a2  +  &2)8. 

4.    (or1  +  ?/3)4.  5.    (a*  —  64)5.  6.    (x~2  +  y?)6. 

7.    (a?_2/t)4.  8.    (a~3 -1- &-1)5.  9.    (w~*-n*)6. 

10.  fc-»y;      11.  («_»y. 

a?y  \6     a/ 


13.    (a-5)6.  14.    (2x  +  3?/)5.  15.    (4  a2  - 

16.      ar'- 


20.     -.  21. 


22.  (V-2+2)7.  23.  (-/a+/&)8.  24.  (a-V~ 
25.  (afr-2-6V)9.  26.  (aj2  -V~^9-  27.  a2&  +  &-3 
28.  ^  +  --14-  29- 


Simplify  each  of  the  following  expressions  : 

so.    i  +    - 


Write  the  expansion  of  each  of  the  following  powers : 
32.    (l—x  +  or)3.  33.    (2  —  3 x  +  or2)4. 

34.    (1  +  a*  -  a-2)3.  35.    (1  -  a;  V2  -h  «V3)4- 

Write  the 

36.   3d  term  of  (a  +  b)15.  37.    5th  term  of  (a  —  6)16. 

38.    6th  term  of  (a&  +  bfy5.         39.    7th  term  of  (an  -  a~n)H. 

40.    6th  term  of  ( ' -tym — E^jV.  41.    15th  term  of  fa3  +  - 

V         vm  /  V 

42.    12th  term  of  (x  -  V~  x)20.    43.   9th  term  of  (^/x  - 

44.  Write  the  middle  term  of  (x  -^/x  —  I)4. 

45.  Write  the  middle  terms  of  (a5  -f  a;^)9. 


CHAPTER   XXIII. 

VARIABLES   AND   LIMITS. 
VARIABLES. 

1.  A  Variable  is  a  number  that  may  have  a  series  of  different 
values  in  the  same  investigation  or  problem. 

A  Constant  is  a  number  that  has  a  fixed  value  in  an  investi- 
gation or  problem. 

Thus,  if  d  stand  for  the  distance  a  body  has  fallen  from  rest 
in  s  seconds,  it  has  been  shown  by  experiment  that 

d  =  16  s2. 

As  the  body  falls,  the  distance  d  and  the  time  s  are  variables, 
and  16  is  a  constant. 

Again,  time  measured  from  a  past  date  is  a  variable,  while 
time  measured  between  two  fixed  dates  is  a  constant. 

2.  The  constants  in  a  mathematical  investigation  are,  as  a 
rule,  general  numbers,  and  are  represented  by  the  first  letters 
of  the  alphabet,  a,  6,  c,  etc. ;  variables  are  usually  represented 
by  the  last  letters,  x,  y,  z,  etc. 

LIMITS. 

3.  When  the  difference  between  a  variable  and  a  constant 
may  become  and  remain  less  than  any  assigned  positive  num- 
ber, however  small,  the   constant  is  called  the  Limit  of  the 
variable. 

Let  the  point  P  move  from  A  toward  B  (Fig.  1)  in  the  fol- 
lowing way :  First  to  Plf  one-half  of  the  distance  from  A  to  B\ 
next  from  Pl  to  P2,  one-half  of  the  distance  from  Pl  to  B\ 

348 


1-5]  LIMITS.  349 

then  from  P2  to  P3,  one-half  of  the  distance  from  P2  to  B  ;  and 
so  on. 


FIG.  1. 

Evidently,  as  P  thus  moves  from  A  to  B,  its  variable  dis- 
tance from  A  becomes  more  and  more  nearly  equal  to  AB,  and 
the  difference  between  AP  and  AB  can  be  made  less  than  any 
assigned  distance,  however  small,  by  continuing  indefinitely 
the  motion  of  P.  Therefore  AB  is  the  limit  of  the  variable  AP. 

If  we  call  the  distance  from  A  to  B  unity,  we  have 


Hence, 
AP1  +  PXP2  +  P*P8  +  AA  +  -  =  i  +  J  4-  i  +  TV  +  -• 

But,  by  Ch.  XXI,  Art.  25,  the  variable  sum  of  the  series  on 
the  right  approaches  1  as  a  limit.     That  is, 

limit  of  (AP,  +  PjP2  +  P2P3  -f  AA  +•••)  =  AB. 

4.  It  follows  from  the  definition  of  a  limit  that  the  variable 
may  be  always  greater,  or  always  less,  or  sometimes  greater 
and  sometimes  less  than  its  limit. 

Thus,  by  Ch.  XXI,  Art.  25,  we  have 

limit  (l-i-i-i-..-)  =  0,  (1) 

limit  (l  +  i  +  J  +  |+..  .)  =  2,  (2) 

limit  (l_£  +  j-£+.  ..)  =  }.  (3) 

And  in  (1),     S1  =  1,  S2  =  i,  S3  =  i,  ^4  =    i,  -  ;              (4) 

in  (2),     ^  =  1,  ^2  =  |,  £3  =  J,  £4  =  V-,  -  ;              (5) 

in  (3),      ^  =  1,  S2  =  I,  ^  =  %,  S*=    *,  --               (6) 

Evidently  the  variable  in  each  of  these  examples  is  the  sum, 
which  changes  as  the  number  of  terms  increases. 

5.  The  symbol,  =,  read  approaches  as  a  limit,  or   simply 
approaches,  is  placed  between  a  variable  and  its  limit. 


350  ALGEBRA.  [Cn.  XXIII 

The  word  limit  may  be  abbreviated  to  lim. 

Thus,  ~?j  (1  —  x)  =.0,  read  the  limit  ofl  —  x,asx  approaches 

1,  is  0. 

Infinites  and  Infinitesimals. 

6.  The  following  considerations  lead  to  important  mathemati- 
cal concepts : 

The  fractions 

o  o  o  o 

_  _  20  •     ^    _  200  *  _  2000  *  _  20000  •  etc 

.1' "        '   .01' "          '  .001'  '  .0001' 

are  particular  values  of  the  fraction  -,  in  which  the  denomina- 
tor x  is  assumed  to  be  a  variable.  It  is  evident  that  the  value 
of  this  fraction  can  be  made  greater  than  any  assigned  number, 
however  great,  by  taking  its  denominator  sufficiently  small. 

A  variable  which  can  become  and  remain  numerically  greater 
than  any  assigned  positive  number,  however  great,  is  called  an 
Infinite  Number,  or  simply  an  Infinite. 

An  infinite  variable  is  denoted  by  the  symbol  GO. 

7.  The  numbers,  variables  and  constants,  which  have  been 
hitherto  used  in  this  book  are,  for  the  sake  of  distinction,  called 
Finite  Numbers. 

8.  The  fractions 

999  9 

±  =  .2  ;  -JL,  =  .02 ;  — — ,  =  .002 ;  — — ,  =  .0002 ;  etc. 
10      100      1000        10000 


?& 

are  also  particular  values  of  the  fraction  — ,  in  which,  as  above, 
the  denominator  x  is  assumed  to  be  a  variable.  It  is  evident 
that  the  value  of  the  fraction  -  can  also  be  made  less  than  any 

assigned  number,  however  small,  by  taking  the  denominator 
sufficiently  great. 

A  variable  which  can  become  and  remain  numerically  less 
than  any  assigned  positive  number,  however  small,  is  called  an 
Infinitesimal. 


5-12]  LIMITS.  351 

No  symbol  by  which  to  denote  an  infinitesimal  variable  has 
been  generally  adopted. 

It  follows  from  the  definition  that  the  limit  of  an  infinitesi- 
mal is  0. 

9.  It  is  important  to  keep  in  mind  that  both  infinites  and 
infinitesimals  are  variables.     Their  definitions  imply  that  fixed 
values  cannot  be  assigned  to  them. 

An  infinitesimal  should  therefore  not  be  confused  with  0, 
which  is  the  constant  difference  between  any  two  equal  numbers. 

10.  The  statement,  x  becomes  infinite,  or  x  increases  numeri- 
cally beyond  any  assigned  positive  number,  however  great,  is  fre- 
quently abbreviated  by  the  expression,  x  =  oo. 

11.  The  conclusions  reached  in  Arts.  6  and  8  can  now  be 
restated  thus : 

(i.)  If  the  numerator  of  a  fraction  remain  finite  and  not  0,  and 
the  denominator  become  infinite,  the  value  of  the  fraction  will 
become  infinite;  or  stated  symbolically, 

n  . 

-  =  oo,  as  x  =  0, 

wherein  n  is  finite  and  not  0. 

(ii.)  If  the  numerator  of  a  fraction  remain  finite  and  not  0,  and 
the  denominator  become  infinite,  the  value  of  the  fraction  will 
approach  0 ;  or  stated  symbolically, 

-  =  0,  as  x  =  co, 

wherein  n  is  finite  and  not  0. 

Observe  that  these  principles  hold  not  only  when  n  is  a 
constant,  not  0,  but  also  when  n  is  a  variable,  provided  it 
does  not  become  infinite. 

12.  The  difference  between  a  variable  and  its  limit  is  evidently 
an  infinitesimal;  that  is, 

if  Urn  x  =  a,  then  lim  (x  —  a)  =  0. 


352  ALGEBRA.  [Cn.  XXIII 

13.  If  the  limit  of  a  variable  be  0,  the  limit  of  the  product  of 
the  variable  and  any  finite  number  is  0  ;  that  is, 

if  lim  x  =  Q,  and  a  be  any  finite  number,  lim  ax  =  0. 
Let  fc  be  any  number,  however  small.     Then  x  can  be  made 

less  numerically  than  -  and,  therefore,  ax  less  than  k.  Hence, 
lim  ax  =  0. 

Indeterminate  Fractions. 

14.  It  follows  from  the  definition  of  a  fraction  that  -  is  a 
number  which  multiplied  by  0  gives  0.     But  any  finite  number 

multiplied  by  0  gives  0,  or  0  n  —  0.  Consequently  -  may  denote 
any  number  whatever. 

For  this  reason,  such  a  fraction  is  called  an  Indeterminate 
Fraction. 

15.  The  fraction  —  —  -  becomes  -  when  x  =  3,  and  has  no 

x  —  o  0 

definite  value.  But  as  long  as  x  =£  3,  however  little  it  may 
differ  from  3,  we  may  perform  the  indicated  division.  We 
therefore  have 

9  =  x  +  3,  when  x  =£  3. 


x  —  3 


Now  since  the  limit  of  the  fraction  depends  upon  values  of 
x  which  differ  from  3,  however  little,  we  have 


Although  the  given  fraction  is  indeterminate,  it  is  clearly 
desirable  that  it  shall  have  a  definite  value.  We  therefore 

y2    _     Q 

assign  to  --  ^-  the  value  6,  when  x  =  3. 

X  —  O 

That  is,  we  define  an  indeterminate  fraction  to  be  the  limit 
of  the  fraction  as  the  variable  approaches  that  value  which 
renders  it  indeterminate.  In  this  way  we  may  obtain  a  definite 
value  when  the  fraction  involves  but  one  variable. 


13-16]  LIMITS.  353 

EXERCISES   I. 
Find  the  limiting  values  of  the  following  fractions  : 


. 

3.2  —  8  a?  -f  15 

/v.2        q  xv,    i    o 

2.     ^~  g  J^,  when  x  =  2. 


2a52  —  a?—  -2       , 
5.    -  —  ,  when  X  =  L. 


6.       l£,  when  aj  =  ! 
7. 


058-30;  +  2 

8.    ^  ~  1,  when  aj  =  0. 
ax  —  1 

Indeterminate  Solutions. 

16.  The  preceding  principles  may  be  further  illustrated  by 
examining  the  infinite  and  indeterminate  solutions  of  certain 
problems. 

Pr.  A  merchant  buys  four  pieces  of  goods.  In  the  second 
piece  there  are  3  yards  less  than  in  the  first,  in  the  third  7 
yards  less  than  in  the  first,  and  in  the  fourth  10  yards  less  than 
in  the  first.  The  number  of  yards  in  the  first  and  fourth  is 
equal  to  the  number  of  yards  in  the  second  and  third.  How 
many  yards  are  there  in  the  first  piece  ? 

Let  x  stand  for  the  number  of  yards  in  the  first  piece  ;  then 
the  number  of  yards  in  the  second  piece  is  x  —  3  ;  in  the  third 
piece,  x  —  1  ;  in  the  fourth  piece,  x  —  10.  Therefore,  by  the 
condition  of  the  problem,  we  have 

x  +  (a;  -  10)  =  (a;  -  3)  +  (aj  -  7),  or  2  x  -  10  =  2  x  -  10. 


354  ALGEBRA.  [Cn.  XXIII 

This  equation  is  an  identity,  and  is  therefore  satisfied  by  any 
finite  value  of  x. 

If  it  be  solved  in  the  usual  way,  we  obtain 

(2  -  2)a;  =  10  -  10,  or  z=^ 


That  is,  the  conditions  of  the  problem  will  be  satisfied  by 
any  number  of  yards  in  the  first  piece. 

Infinite  Solutions. 

17.  Pr.  A  cistern  has  three  pipes.  Through  the  first  it  can 
be  filled  in  24  minutes;  through  the  second  in  36  minutes; 
through  the  third  it  can  be  emptied  in  a  minutes.  In  what 
time  will  the  cistern  be  filled  if  all  the  pipes  be  opened  at  the 
same  time  ? 

Let  x  stand  for  the  number  of  minutes  after  which  the 
cistern  will  be  filled.  In  one  minute  -fa  of  its  capacity  enters 
through  the  first  pipe,  and  hence  in  x  minutes  -ffx  of  its 
capacity  enters.  For  a  similar  reason,  -f^x  of  its  capacity 
enters  through  the  second  pipe  in  x  minutes  ;  and  in  the  same 

time  -  x  of  its  capacity  is  discharged  through  the  third  pipe. 

a 
Therefore,  after  x  minutes  there  is  in  the  cistern 


of  its  capacity.     But  by  the  condition  of  the  problem,  that  the 
cistern  is  then  filled,  we  have 


whence  x  = 


If  we  now  let  a  approach  -f^  then  x  becomes  infinite. 

This  result  would  mean  that  the  cistern  would  never  be 
filled.  This  is  also  evident  from  the  data  of  the  problem,  since 
the  third  pipe  in  a  given  time  would  discharge  from  the 
cistern  as  much  as  would  enter  it  through  the  other  pipes. 


16-18]  LIMITS.  355 


The  Problem  of  the  Couriers. 

18.  Pr.  Two  couriers  are  travelling  along  a  road  in  the 
direction  from  M  to  N\  one  courier  at  the  rate  of  ml  miles  an 
hour,  the  other  at  the  rate  of  m2  miles  an  hour.  The  former 

A  B 

M  C3  C2  C\  If 

FIG.  2. 

is  seen  at  the  station  A  at  noon,  and  the  other  is  seen  h  hours 
later  at  the  station  B,  which  is  d  miles  from  A  in  the  direction 
in  which  the  couriers  are  travelling.  Where  do  the  couriers 
meet  ? 

Assume  that  they  meet  to  the  right  of  B  at  a  point  Ci,  and 
let  x  stand  for  the  number  of  miles  from  B  to  the  place  of 
meeting  C\  (Fig,  2). 

The  first  courier,  moving  at  the  rate  of  m1  miles  an  hour, 

(  1      I      w 

travels  d  -f  x  miles,  from  A  to  Clf  in  -  hours  ;  the  second 

wti 

courier,  moving  at  the  rate  of  ra2  miles  an  hour,  travels  x  miles, 

n,* 

from  B  to  d,  in  —  hours.     By  the  condition  of  the  problem 
m2 

it  is  evident  that,  if  the  place  of  meeting  is  to  the  right  of  B, 
the  number  of  hours  it  takes  the  first  courier  to  travel  from 
A  to  Ci  exceeds  by  h  the  number  of  hours  it  takes  the  second 
courier  to  travel  from  B  to  Ci.     We  therefore  have 
d  +  x      x 


ml 


—  d} 

'- 


whence  x  = 

m2  —  m1  m2  —  ml 

(i.)  A  Positive  Result.  —  The  result  will  be  positive  either 
when  hml  >  d  and  ra2  >  mlt  or  when  hmt  <  d  and  m2  <  m^  A 
positive  result  means  that  the  problem  is  possible  with  the 
assumption  made;  i.e.,  that  the  couriers  meet  at  a  point  to 
the  right  of  B. 

(ii.)  A  Negative  Result.  —  The  result  will  be  negative  either 
when  hml  >  d  and  m2  <  m1}  or  when  hm^  <  d  and  m2  >  ra^  Such 


ALGEBRA.  [Cn.  XXIII 

a,  result  shows  that  the  assumption  that  the  couriers  meet  to 
the  right  of  B  is  untenable,  since,  as  we  have  seen,  in  that 
case  the  result  is  positive. 

That  under  the  assumed  conditions  the  couriers  can  meet 
only  at  some  point  to  the  left  of  B  can  also  be  inferred  from 
the  following  considerations,  which  are  independent  of  the 
negative  result :  If  lim^  >  d,  the  first  courier  has  passed  B  when 
the  second  courier  is  seen  at  that  station ;  that  is,  the  second 
courier  is  behind  the  first  at  that  time.  And  since  also 
w2  <  nil)  the  first  courier  is  travelling  the  faster,  and  must 
therefore  have  overtaken  the  second,  and  at  some  point  to  the 
left  of  B. 

On  the  other  hand,  if  hm^  >  d,  the  first  courier  has  not  yet 
reached  B  when  the  second  is  seen  at  that  station ;  that  is,  the 
first  courier  is  behind  the  second  at  that  time.  And  since  also 
w2  >  m1?  the  second  courier  is  travelling  the  faster,  and  must 
therefore  have  overtaken  the  first,  at  some  point  to  the  left  of 
B.  Similar  reasoning  could  have  been  applied  in  (i.). 

(iii.)  A  Zero  Result.  —  A  zero  result  is  obtained  when 
/iWj  =  d,  and  m2  is  not  equal  to  ?% ;  that  is,  the  meeting  takes 
place  at  B.  This  is  also  evident  from  the  assumed  conditions. 
For  the  first  courier  reaches  B  h  hours  after  he  was  seen  at  A ; 
and  since  the  second  courier  is  seen  at  B  h  hours  after  the 
first  was  seen  at  A<  the  meeting  must  take  place  at  B. 

(iv.)  Indeterminate  Result.  —  An  indeterminate  result  is  ob- 
tained if  hnii  =  d,  and  m2  =  m^  In  this  case  every  point  of 
the  road  can  be  regarded  as  their  place  of  meeting.  For  the 
first  courier  evidently  reaches  B  at  the  time  at  which  the 
second  courier  is  seen  at  that  station  ;  and  since  they  are  trav- 
elling at  the  same  rate,  they  must  be  together  all  the  time. 
The  problem  under  these  conditions  becomes  indeterminate. 

(v.)  An  Infinite  Result.  —  An  infinite  result  is  obtained  when 
hml  ^  d,  and  m2  =  ?%.  In  this  case  a  meeting  of  the  couriers 
is  impossible,  since  both  travel  at  the  same  rate,  and  when  the 
second  is  seen  at  B  the  first  either  has  not  yet  reached  B  or 
has  already  passed  that  station. 


18]  LIMITS.  357 

An  infinite  result  also  means  that  the  more  nearly  equal 
mi  and  m.2  are,  the  further  removed  is  the  place  of  meeting. 

EXERCISES    II. 

Solve  the  following  problems,  and  interpret  the  results : 

1.  In  a  number  of  two  digits,  the  digit  in  the  tens7  place 
exceeds   the  digit  in  the  units'  place  by  5.     If  the  digits  be 
interchanged,  the  resulting  number  will  be  less  than  the  origi- 
nal number  by  45.     What  is  the  number  ? 

2.  The  sum  of  the  first  and  third  of  three  consecutive  even 
numbers  is  equal  to  twice  the  second.    What  are  the  numbers  ? 

3.  A  father  is  26  years  older  than  his  son,  and  the  sum  of 
their  ages  is  26  years  less  than  twice  the  father's  age.     How 
old  is  the  son  ? 

4.  In  a  number  of  two  digits,  the  digit  in  the  units'  place 
exceeds  the  digit  in  the  tens'  place  by  4.     If  the  sum  of  the 
digits  be  divided  by  2,  the  quotient  will  be  less  than  the  first 
digit  by  2.     What  is  the  number? 

Discuss  the  solutions  of  the  following  general  problems : 

5.  What  number,  added  to  the  denominators  of  the  fractions 

-  and  -,  will  make  the  resulting  fractions  equal  ? 
b          d 

6.  Having  two  kinds  of  wine  worth  a  and  b  dollars  a  gallon, 
respectively,  how  many  gallons  of  each  kind  must  be  taken  to 
make  a  mixture  of  n  gallons  worth  c  dollars  a  gallon  ? 

7.  Two  couriers,  A  and  B,  start  at  the  same  time  from  two 
stations,  distant  d  miles  from  each  other,  and  travel  in  the 
same  direction.     A  travels  n  times  as  fast  as  B.     Where  will 
A  overtake  B  ? 


CHAPTER   XXIV. 

UNDETERMINED   COEFFICIENTS. 
CONVERGENT  AND   DIVERGENT  SERIES. 
1.   The  infinite  series 


is  a  decreasing  geometrical  progression,  whose  ratio  is  -|  .  It 
follows  from  Ch.  XXI.,  Art.  26,  that  the  sum  of  this  series 
approaches  a  definite  finite  value  as  the  number  of  terms  is 
indefinitely  increased. 

Let  Sn  =  1  +  |  +  |  +  Jy  H  ----  to  n  terms. 

Then,  by  Ch.  XXI.,  Art.  26, 


as  n  increases  indefinitely. 

By  actual  computation,  we  obtain 

S,  =  1,  S,  =  If,  £3  =  2i   S4  =  2^,  etc. 

These  sums  approach  3  more  and  more  nearly,  as  more  and 
more  terms  are  included. 

This  infinite  series  may  therefore  be  regarded  as  having  the 
finite  sum  3. 

But  the  sum  of  the  series 


increases  beyond  any  finite  number,  as  the  number  of  terms 
increases  indefinitely. 

2.   The  examples  of  the  preceding  article  illustrate  the  fol- 
lowing definitions  : 

358 


1-4]          CONVERGENT   AND   DIVERGENT   SERIES.          359 

Any  infinite  series  is  said  to  be  Convergent,  when  the  sum  of 
the  first  n  terms  approaches  a  definite  finite  limit,  as  n  increases 
indefinitely. 

An  infinite  series  is  said  to  be  Divergent  when  the  sum  of  the 
first  n  terms  increases  numerically  beyond  any  assigned  num- 
ber, however  great,  as  n  increases  indefinitely. 

3.  It  was  shown  in  Ch.  XXI.,  Art.  26,  that,  when  r  <  1,  the 
sum  of  the  series 

a  +  ar  -f-  cut*  +  •  •  • 

approaches  the  definite  finite  value  —  —  —    as  the  number  of 

1  —  r 
terms  is  indefinitely  increased. 

Therefore,  any  decreasing  geometrical  progression  is  a  con- 
vergent series. 

4.  Infinite  series  arise  in  connection  with  many  mathematical 
operations.     Thus,  for  example,  if  the  division  of  1  by  1  —  x 
be  continued  indefinitely,  we  obtain  as  a  quotient  the  infinite 
series 


When  x  is  numerically  less  than  1,  this  series  is  a  decreasing 
geometrical  progression,  as  in  Art.  1.  Therefore,  by  the  preced- 
ing article  it  is  convergent. 

When  x  =  1,  the  series  becomes 


and  is  evidently  divergent. 
When  x  —  —  1,  we  have 

1  _  1  +  1  _!  +  .... 

The  sum  of  n  terms  of  this  series  is  -1-1  or  —  1,  according 
as  n  is  odd  or  even.  The  series  is  said  to  oscillate  and  is  neither 
convergent  nor  divergent. 

When  x  is  numerically  greater  than  1,  we  have,  by  Ch.  XXI., 
Art,  22  (II.), 

o        1  —  aJ" 


360  ALGEBRA.  [Cn.  XXIV 

By  taking  n  sufficiently  great  this  expression  can  be  made  to 
exceed  numerically  any  number,  however  great. 
Therefore  the  series  is  divergent. 
Thus,  when  x  =  2,  the  series  becomes 


The  sum  of  this  series  can  evidently  be  made  greater  than 
any  assigned  number,  however  great.      But  when  x  =  2,  the 

value  of  the  fraction  -  is  —    —  ,  =  —  1. 
We  therefore  conclude  that  the  infinite  series 


approaches  in  value  the  fraction  -      -  for  all  values  of  x  be- 

1  —  x 

tween  —  1  and  +  1.  Conversely,  we  may  look  upon  the  series 
as  the  expansion  of  the  fraction  for  all  values  of  x  between 
these  limits,  but  for  no  other  values  of  x. 

In  general,  an  infinite  series,  no  matter  how  obtained  from  a 
given  expression,  can  be  regarded  as  the  expansion  of  that 
expression  only  when  the  series  is  convergent. 

This  fact  should  be  kept  in  mind,  without  further  emphasis, 
in  all  the  expansions  that  we  shall  derive  in  this  chapter. 


5.  If  an  infinite  series  a0  +  a^x  +  a>$?  +  a$?  +  •  ••  be  con- 
vergent for  values  of  x  greater  than  0,  the  sum  of  the  series 
approaches  a0,  as  x  approaches  0. 

Let  a0  +  a^  +  a#?  +  a3or3  -\  ----  =  a0  +  xS» 

wherein  S1  =  al  +  a&  +  a3x2  +  •••. 

Evidently,  if  the  given  series  is  convergent,  that  is,  if 
a0  +  xSi  is  finite,  then  Si  is  finite.  Therefore,  by  Ch.  XXIII.  , 
Art.  13,  xSi  =  0,  when  x  =  0. 

Consequently 

a0  +  aix  +  a$?  +  «  •  «,   —  a0  +  XS,   =  «o>  when  x  =  0. 


6.  If  two  integral  series,  arranged  to  ascending  powers  of  x, 
be  equal  for  all  values  of  x  which  make  them  both  convergent,  the 
coefficients  of  like  powers  of  x  are  equal. 


4-8]  EXPANSION   OF   FRACTIONS.  361 


Let          a0  +  a^x  -f-  a.-p?  -f-  •••  =  bQ  4-  b& 

for  all  values  of  #  which  make  the  two  series  convergent. 

Then  the  sums  of  the  two  series  approach  equal  limits  when 
x  =  0.  But,  by  the  preceding  article,  the  sum  of  the  one  series 
approaches  a0,  that  of  the  other  bQ  ;  consequently  a0  —  60, 

and  «!»  +  0*1?+  "-  =  b]X  +  b$?  +  ••-. 

Since  these  two  series  are  convergent  for  all  values  of  x  for 
which  the  original  series  are  convergent,  they  are  equal  for 
values  of  x  other  than  zero,  and  the  last  equation  may  be 
divided  by  x. 

Hence 

%  +  <*>&  -f  a$?  H  ----  =  &i  +  b2x  +  bsx2  -\  ----  ; 

and,  as  before,  a^  =  b^ 

and  a^x  +  a^x2  -\  ----  =  b&  +  b$?  +  ~>. 

In  like  manner,  we  can  prove  a2  =  b2,  as  =  b3,  etc. 

7.  Evidently  the  principle  of    the  preceding  article  holds 
with  greater  reason  if  .either  or  both  of  the  series  be  finite, 
i.e.,  have  a  limited  number  of  terms.     There  is,  in  this  case, 
no  question  of  convergence  of  the  finite  series.      The  series 
must  be  equal  for  all  values  of  x,  if  they  be  both  finite  ;  or, 
if  one  be  infinite,  for  all  values  of  x  which  make  that  series 
convergent. 

EXPANSIONS   OF   RATIONAL   FRACTIONS. 

8.  We  shall  now  give  a  method  of  expanding  a  fraction  in 
an  infinite  series,  without  performing  the  actual  division. 

Ex.  1.    Expand  2  ~  x     •> 

1  +  x  —  x2 

in  a  series,  to  ascending  powers  of  x. 

We  equate  the  fraction  to  a  series  of  the  required  form,  in 
which  the  coefficients  of  the  different  powers  of  x  are  un- 
known, or  undetermined. 

Assume  —  2  ~  x   0  =  A  +  Bx  +  Cx2  +  Dx*  +  Ex4  •••, 

1  +  x  —  x* 

whence  A,  B,  (7,  D,  E,  •••  are  constants  to  be  determined. 


362 


ALGEBRA. 


[Cn.  XXIV 


€learing  the  equation  of  fractions,  we  obtain 


2-x=A+B 
A 


x+C 
+  B 


-B 


-  C 


In  this  work  the  powers  of  x  in  the  terms  of  the  second 
and  third  partial  products  are'  omitted,  it  being  understood 
that  the  letters  remaining  are  the  coefficients  of  the  powers 
of  x  just  above  in  the  first  partial  product. 

Thus  the  coefficient  of  x  is  A  +  B,  etc. 

The  series  on  the  right  is  infinite  ;  that  on  the  left  may  be 
regarded  as  an  infinite  series  with  zero  coefficients  of  all 
powers  of  x  higher  than  the  first.  By  Art.  6,  we  have 

B  +  A  =  ^-1,  whence  B^  —  3; 

C+B-A  =  0,  whence  (7-5; 

C-B  =  Q,  whence  Z>  =  -8; 

D-C=Q,  whence  E  =  13; 


etc., 


etc. 


Hence,   substituting   these  values   of  A,  B,  (7,  D, 
assumed  series,  we  have 

2-x 


in  the 


1+x-x2 

We  can  assume  that  this  series  is  equivalent  to  the  fraction 
only  when  x  has  such  values  as  make  it  convergent. 

Let  the  student  compare  tliis  result  with  that  obtained  by 
division.  In  fact,  the  latter  method  of  expanding  a  fraction  is 
to  be  preferred  when  only  a  few  terms  are  wanted.  But  the 
successive  coefficients,  after  a  certain  stage,  may  be  computed 
with  great  facility  by  the  method  of  undetermined  coefficients. 
A  moment's  inspection  of  the  preceding  work  will  convince 
the  student  that  the  coefficient  Z>,  and  all  which  follow  it,  are 
each  connected  with  the  two  immediately  preceding  coefficients 
by  a  definite  relation.  Thus, 

D+C-  B  =  Q,  E  +  D-  C=0,  F+  E-D=  0,  etc. 


8]  EXPANSION  OF   FRACTIONS.  363 

In  assuming  as  the  expansion  of  a  rational  fraction  an 
infinite  series  of  ascending  powers  of  x,  it  is  usually  necessary 
tirst  to  determine  with,  what  power  the  series  should  com- 
mence. This  is  done  by  division,  when  both  numerator  and 
denominator  are  arranged  to  ascending  powers  of  x.  In  fact, 
this  step  also  determines  completely  the  first  term  of  the 
series. 

Ex.2.   Expand  f 


in  a  series  to  ascending  powers  of  x. 

The  first  term  in  the  expansion,  obtained  by  division,  is 
evidently  ^x~2. 

We  therefore  assume 

— ~  x  =  i  x~2  -\-Bx~1  4-  C  -f-  Dx  - 
Clearing  of  fractions,  we  obtain 


B 


3C 
-B 


-C 


By  Art.  6,  we  have 

1  =  1,     3B  —  ^  =  —  1,  whence  B  =  — 

3C-B=       0,  whence  C=  — 

3  D  —  C  =       0,  whence  D=  — 

etc.,  etc. 

Hence' 


EXERCISES   I. 

Expand  the  following  fractions  in  series,  to  ascending  powers 
of  x,  to  four  terms  : 

1          1  2          3  3         6 

l—2x  '   l  +  3x  '  3-x 

—  2 


4  .  5 

'   l-x  ' 


7.  .  a          ~  9. 


364 
10. 


x  _  3* 


11. 


ALGEBRA. 
2-x 


[Cii.  XXIV 


13.    2+*-3<         14.    ^=-3-^ 


3  -  x  +  3  a? 


12. 


15. 


EXPANSION   OF   SURDS. 

9.   Ex.   Expand         V(1-^2  +  2^ 
in  a  series,  to  ascending  powers  of  x.     Assume 
^/(l  _  or2  +  2  a3)  =  1  +  Bx  +  Os2  +  Dx3  + 
Squaring  both  sides  of  the  equation,  we  have 


2BC 


+  2BD 


Equating  coefficients,   1  =  1. 

25  =  0,  whence  5  =  0; 

2  C  +  B2  =  -  1,  whence  C  =  -  J ; 

2  Z)  +  2  5(7  =  2,  whence  Z)  =  +  1 ; 

2 .E  +  2 BD  +  C*  =  0,  whence  E  =  —%\  etc. 

Hence    V(l  -  *?  +  2a^)=  1  - 


EXERCISES  II. 

Expand  the  following  expressions  in  series,  to  ascending 
powers  of  x,  to  four  terms  : 

i.  v(1  +  a)-  2-  V(«2-2^)-         3- 

4.    ^(4  -  2  a;  +  a2).      5.    V(5+3a;+9^)-      6- 

PARTIAL  FRACTIONS. 

10.  It  is  frequently  desirable  to  separate  a  rational  alge- 
braical fraction  into  the  simpler  (partial)  fractions  of  which  it 
is  the  algebraical  sum. 

2x  1  1 


E.g., 


1-x2     1-x 


8-9]  PARTIAL   FRACTIONS.  365 

The  process  of  separating  a  given  fraction  into  its  partial 
fractions  is,  therefore,  the  converse  of  addition  (including  sub- 
traction) of  fractions ;  and  this  fact  must  guide  us  in  assuming 
the  forms  of  the  partial  fractions. 

We  shall  also  assume  that  the  degree  of  the  numerator  is  at 
least  one  less  than  that  of  the  denominator.  A  fraction  whose 
numerator  is  of  a  degree  equal  to  or  greater  than  that  of  its 
denominator  can  be  first  reduced  by  division  to  the  sum  of  an 
integral  expression  and  a  fraction  satisfying  the  above  condi- 
tion. The  latter  fraction  will  then  be  decomposed. 

The  denominators  of  the  partial  fractions  can  be  definitely 
assumed.  For  they  are  evidently. those  factors  whose  lowest 
common  multiple  is  the  denominator  of  the  given  fraction. 
But  there  is  one  case  of  doubt ;  namely,  when  a  prime  factor  is 
repeated  in  the  denominator  of  the  given  fraction. 

E.g., 

6-2X2       =     3         2_          1 

(1  -  x)2  (1  +  x)      1  -  x     (1  -  x)2     1  +  x ' 

3  +  x2  2  1 


We  could  not  have  decided,  in  advance,  whether  either  of 
the  two  given  fractions  is  the  sum  of  two  or  of  three  partial 
fractions.  There  must  necessarily  be  a  partial  fraction  having 
(1  —  x)2  as  a  denominator,  since,  otherwise,  the  L.  C.  M.  of  the 
denominators  would  not  contain  the  prime  factor  1  —  x  to 
the  second  power.  But  it  cannot  be  determined,  in  advance, 
whether  there  is  a  partial  fraction  having  1  —  x  as  a  denomi- 
nator. 

In  such  cases,  therefore,  it  is  advisable  to  make  provision 
for  all  possible  partial  fractions  by  assuming  as  denominators 
all  repeated  factors  to  the  first  power,  second  power,  etc. 

The  numerators  of  partial  fractions  thereby  assumed,  which 
should  not  have  been  included,  will  acquire  the  value  zero 
from  the  subsequent  work,  so  that  those  fractions  drop  out  of 
the  result. 


366  ALGEBRA.  [Cn.  XXIV 

The  numerators  of  the  partial  fractions  must  be  assumed 
with  undetermined  coefficients.  Since  the  numerator  of  the 
given  fraction  is,  by  the  hypothesis,  of  degree  at  least  one  less 
than  the  denominator,  the  same  must  be  true  of  each  partial 
fraction.  We  therefore  assume,  for  each  numerator,  a  complete 
rational  integral  expression  with  undetermined  coefficients  of 
degree  one  lower  than  the  corresponding  denominator. 

If  any  term  in  the  assumed  form  of  the  numerator  should 
not  have  been  included,  its  coefficient  will  prove  to  be  zero. 

An  exception  to  this  principle  occurs  when  the  denominator 
of  the  partial  fraction  is  the  second  or  higher  power  of  a  prime 
factor,  as,  (1  —  x)2.  In  that  case  the  numerator  is  assumed  as 
it  would  be  according  to  the  above  principle  if  the  prime  factor 
occurred  to  the  first  power  only. 

We  may  briefly  restate  the  above  principles : 

Separate  the  denominator  of  the  given  fraction  into  its  prime 
factors.  Assume  as  the  denominator  of  a  partial  fraction  each 
prime  factor ;  in  particular,  when  a  prime  factor  enters  to  the  nth 
power,  assume  that  factor  to  the  first  power,  second  power,  and  so 
on,  to  the  nth  power,  as  a  denominator. 

Assume  for  each  numerator  a  rational  integral  expression,  with 
undetermined  coefficients,  of  degree  one  lower  than  the  prime 
factor  in  the  corresponding  denominator. 

Let  us  first  decompose  the  two  fractions  which  we  have  used 
to  illustrate  the  theory. 


(1  -  x)2(l  +  x)      1  -  x     (1  -  xf     1  +  x 

Since  the  prime  factor  in  the  denominator  of  each  partial 
fraction  is  of  the  first  degree,  each  numerator  is  assumed  to  be 
of  the  zeroth  degree. 

Clearing  the  equation  of  fractions,  we  have 

6  -  2  or2  =  A(l  -  x)(l  +  x)  +  B(l  +  »)  +  C(l  -  x)2 


9]  PARTIAL  FRACTIONS.  367 

Since  this  equation  must  be  true  for  all  values  of  x,  we  have 

B-2C=     0,  1  Whence^l  =  3,  5  =  2,  O^l. 

A  +  B+C=      6.  j 

Ex.2.     „     3  +  f°.       =-A_  +  7-A_H_     # 


(1  -  x)\l  +  x)      1  -  x     (1  -  x)2     1  +  a 

The  forms  of  the  partial  fractions  are  assumed  the  same  as 
in  Ex.  1.  '  We  have 

3  +  x*  =  (-A+C)x*  +  (B-2C)x  +  A  +  B+C, 
and  then         —  A  +  C  =  1, 


5-20=0, 


Therefore  3  -f 


Whence  ^1  =  0,  B  =  2}  (7  =  1. 


When  the  factors  of  the  denominator  of  the  given  fraction 
are  of  the  first  degree,  as  in  Exs.  1  and  2,  the  work  may  be 
shortened. 

Begin  with  the  equation 

6  -  2  tf  =  A(l  -  x)  (1  +  x)  +  B (1  +  x)  +  C (1  -  x)2, 

of  Ex.  1.     Since  this  equation  is  true  for  all  values  of  x,  we 
may  substitute  in  it  for  x  any  value  we  please.     Let  us  take 
such  a  value  as  will  make  one  of  the  prime  factors  zero. 
Substituting  1  for  x,  we  obtain 

4  =  25,  whence  B  =  2. 
Next,  letting  x  =  —  1,  we  have 

4  =  40,  whence  O=l. 

There  is  no  other  value  of  x  which  will  make  a  prime  factor 
zero,  but  any  other  value,  the  smaller  the  better,  will  give  an 
equation  in  which  we  may  substitute  the  values  of  B  and  Q 
already  obtained.  * 


ALGEBRA.  [Cn.  XXIV 


Letting  x  =  0,  we  obtain 

6  =  A  +  B  +  (7,  whence  A  =  3. 
The  same  method  can  be  applied  to  Ex.  2. 


Ex  3    a?2  —  a?  +  3=         or*  -#  +  3          =     .4 
a8-!          a-laj'  +  aj  +  l       o?-l 


In  this  example,  the  one  prime  factor  being  of  the  second 
degree,  we  assume  the  corresponding  numerator  to  be  a  com- 
plete linear  expression. 

Clearing  of  fractions,  we  have 

x2  -  x  +  3  =  A  (x2  +  x  +  1)  +  (Bx  +  C)  (aj  -  1)  = 

(A  +  J3)aj*  +  (A  -  B  +  C)x  +  -4  -  C. 
Equating  coefficients  of  like  powers  of  x,  we  obtain 
-4  +  ^  =  1,  A-B+C  =  -1,  A-C  =  3; 
whence,  ^1  =  1,  5-0,  C  =  -  2. 

Or,  we  might  have  used  the  second  method,  beginning  with 

a-2  -  a;  +  3  =  4  (a2  +  a;  +  1)  +  (jB.e  +  0)  (o>  -  1). 
Letting  a;  =  1,  we  obtain 

3  =  3-4,  whence  A  =  l. 

Since  no  other  value  of  x  will  make  a  factor  vanish,  we  take 
any  simple  values.  When  x  =0,  we  have 

3  =  A  -  C,  whence  G  =  -2. 
Finally,  letting  x  =  —  1,  we  have 

5  =  A-2B-2C,  whence  B  =  0. 
*  -  x  +  3          1  2 


Therefore 


x-1     or' 


2-2a  +  4ft2   =Ax  +  B      Cx  +  D         E 

"  /-i  O\  O    s  -I  \  -4          .  •>          I         /-«         .  O\  O       • 


(1  +  x2)2(l  -x)       1  +  x2       (1  +  x2)2     1  -  a; 

The  prime  factors  in  the  denominator  of  the  first  two  partial 
fractions  being  of  the  second  degree,  expressions  of  the  first 
degree  are  assumed  as  numerators. 


9]  PARTIAL  TRACTIONS.  369 

Clearing  of  fractions,  we  have 
2_2#  +  4cc2 

=  (Ax  +  B)  (I  +  x2)  (1  -  a?)  +  (Cx  +  D)  (1  -  x)  +  E  (1  +  z2)2 
-  (  -  A  +  jBJ)  a?*  +  (A  -  B)  x3  +  (  -  ^L  +  £  -  C  +  2  J0)  x2 

+  (A-B+C-U)x+(B  +  D  +  E). 
Equating  coefficients  of  like  powers  of  x,  we  obtain 


whence,         A  =  l,  B=l,  C=-2,  D  =  0,  E  =  l. 

Therefore 


EXERCISES   III. 

Separate  the  following  fractions  into  partial  fractions  : 
1                 6                                                       7 

(a>-2)(l-2aO 
3x  —  1 

"   (5  +  3x)(,  +  4) 
1             l-« 

•    («,  +  3)(*-2) 

^               ^ 

fi                                                        7 

•  T-y 

8 

ar  —  4                               9  —  or 
9    x  -\-  ^  x      1^ 

10           3  a?  +  ^ 

9a^-16 

'    (*2-l)(*-2) 

U<    6(aj«-9)(aj-3) 
13            ,2  +  5^  +  10 

L"    (*  +  !)(*  -I)2 
o  a;  (a;  +  o) 

(^  +  l)(x  +  2)(x  +  3) 
15                        3~X 

1 

1                                       2 

17                      .                                1R                     •> 

^      (.-I)3' 

19      ^  +  1 

'     3?-l                                         '     X*+l 

°0                r                        21                      • 

19>    ?~1 

CHAPTER  XXV. 

THE   BINOMIAL   THEOREM   FOR   ANY  RATIONAL 
EXPONENT. 

1.   From  Ch.  XXII.,  Art.  4,  we  have 


when  n  is  a  positive  integer.  In  this  case,  as  we  have  seen, 
the  series  ends  with  the  n  -f-  1th  term.  But  if  n  be  not  a  posi- 
tive integer,  the  expression  on  the  right  of  (1)  will  continue 
without  end,  since  no  factor  of  the  form  n  —  k  can  reduce  to  0. 
Therefore  the  series  will  have  no  meaning  unless  it  be  con- 
vergent. 

2.  It  is  proved  in  Elements  of  Algebra,  Ch.  XXXI.,  that 
this  series  is  convergent  when  x  lies  between  —  1  and  +  1  ; 
and,  in  Ch.  XXXII.,  that  when  the  series  is  convergent,  it  is 
the  expansion  of  (1  -f  x)n. 

Ex. 


This   infinite   series    can    be    taken    as    the    expansion    of 
(1  -f  a?)^,  =V(1  +ic)  onty  when  x  is  numerically  less  than  1. 

3.   Expansion  of  (a  +  6)".  —  We  have 

(1) 


and  (a  +  ft).  =        i+=i+  (2) 

370 


1-4]  THE   BINOMIAL   THEOREM.  371 

When  b  is  numerically  less  than  a, 


a 
and,  by  (1)  above, 


..   (3) 

In  a  similar  way  it  can  be  shown  that,  when  a  is  numerically 
less  than  6, 

(a  +  b)n  =  bn  +  nb*~la  +  M^~91V-*a*  +  ....  (4) 

Notice  that  when  n  is  a  fraction  or  negative,  formula  (3)  or 
(4)  must  be  used  according  as  a  is  numerically  greater  or  less 
than  6. 

4.   Ex.  1.    Expand  -—  -  -to  four  terms. 

S/(a-462) 

If  we  assume  a  >  4  62,  by  (3),  Art.  3,  we  have 


=  32  54         896  66 

9  a2^/a     81  a?^/a 


If  a  <  4  b2,  we  should  have  used  (4),  Art.  3. 

Any  particular  term  can  be  written  as  in  Ch.  XXII.,  Art.  9. 

/  1  \  —  2 

Ex.  2.   The  6th  term  in  the  expansion  of  f  x  —  —  \     is 

\       °^J 


(-2)  (-3)  (-4)  (-5)  (-6)     2_5 

1.2.3.4.5 


372  ALGEBRA.  [Cn.  XXV 

5.   Extraction  of  Roots  of  Numbers.  —  Ex.     Find  ^/VI  to  four 
decimal  places.     We  have 


=  4  (1  +  .03125  -  .00048  +  .00001  ----  ) 
=  4  x  1.03078  =  4.12312. 
Therefore  -^/17  =  4.1231,  to  four  decimal  places. 


EXERCISES. 

Expand  to  four  terms  : 


1. 

(1  +  a)*. 

2.    (1-x)-1. 

3.     (1  -  £)-3. 

4. 

(1  +  x^. 

5.    (1  +  a)-4. 

6.    (1  —  ?/2)~2. 

7. 

(x*  +  y)-i. 

8.    (x  —  y2)-*. 

9.    (27  +  5z)i 

10. 

(8a3-36)i 

11.    (3  +  2  a)*. 

19        /'^  /-»2          Q  7i3\"3 
A<&.      (  O  Ct    —  O  0   )      . 

13 

1 

14. 

1 

C«2-62)  ^(a3-6) 

v    \ 

Find  the 

16.   4th  term  of  (1-2  z)i          17.   6th  term  of  (1 

18.  5th  term  of  (a;*  -  ar1/)"^. 

19.  8th  term  of  (a3 ^/b  —  2 

20.  fc — 5th  term  of  (1  + 

21.  2  fcth  term  of  T^2  - 


Find  to  four  places  of  decimals  the  values  of : 
22.    ^/5.        23.    ^/27.       24. 


CHAPTER   XXVI. 

LOGARITHMS. 

1.  It  is  proved  in  Elements  of  Algebra,  Ch.  XXXV.,  that  a, 
value  of  x  can  always  be  found  to  satisfy  an  equation  of  the 
form 

10*  =  n, 

wherein  n  is  any  real  positive  number.     E.g.,  when  n  =  10, 
x  =  1,  when  n  =  100,  x  =  2,  when  n  =  1000,  x  =  3,  etc. 

When  n  is  not  an  integral  power  of  10,  the  value  of  x  is  irra- 
tional, and  can  be  expressed  only  approximately.  Thus,  when 
n  =  24,  the  corresponding  value  of  x  has  been  found  to  be 
1.38021*",  to  five  decimal  places;  or 

101.38021-  =  24. 

A  value  of  x  is  called  the  logarithm  of  the  corresponding 
value  of  n,  and  10  is  called  the  base. 

In  general,  a  value  of  x  which  satisfies  the  equation  b*  =  n, 
is  called  the  logarithm  of  n  to  the  base  b. 

E.g.,  since  23  =  8,  3  is  the  logarithm  of  8  to  the  base  2 ;. 
since  102  =  100,  2  is  the  logarithm  of  100  to  the  base  10. 

The  Logarithm  of  a  given  number  n  to  a  given  base  b  is,  there- 
fore, the  exponent  of  the  power  to  which  the  base  b  must  be 
raised  to  produce  the  number  n. 

2.  The  relation  bx  —  a  is  also  written  x  =  Iog6  a,  read  x  is 
the  logarithm  of  a  to  the  base  b.     Thus, 

23  =  8       and  3  =  log28, 
102  =  100  and  2  =  logw100, 

are  equivalent  ways  of  expressing  one  and  the  same  relation. 

373 


374  ALGEBRA.  [Cn.  XXVI 

3.  The   theory  of   logarithms   is   based   upon   the  idea  of 
representing  all  positive  numbers,  in  their  natural  order,  as 
powers  of  one  and  the  same  base. 

Thus,  4,  8,  16,  32,  64,  etc.,  can  all  be  expressed  as  powers  of 
a  common  base  2 ;  as  4  =  22,  8  =  23,  16  =  24,  etc.  Since,  also, 
all  the  numbers  intermediate  between  those  given  above  can 
be  expressed  as  powers  of  2,  the  exponents  of  these  powers 
are  the  logarithms  of  the  corresponding  numbers. 

The  logarithms  of  all  positive  numbers  to  a  given  base  form 
what  is  called  a  System  of  Logarithms.  The  base  is  then  called 
the  base  of  the  system. 

It  follows  from  Art.  1,  that  any  positive  number  except  1 
may  be  taken  as  the  base  of  a  system  of  logarithms. 

EXERCISES   I. 

Express  the  following  relations  in  the  language  of  logarithms : 
1.   52  =  25.          2.    25  =  32.         3.   73  =  343.         4.   37  =  2187. 

Express  the  following  relations  in  terms  of  powers : 
5.  Iog381  =  4.      6.  Iog981  =  2.      7.  Iog464  =  3.      8.  Iog264  =  6. 

Determine  the  values  of  the  following  logarithms  : 
9.   Iog232.  10.   Iog^l28.         11.   Iog2.5.  12.    Iog2.25. 

13.    Iog464.          14.    log^S.  15.   Iog2.125.        16.    Iog5 .04. 

To  the  base  16,  what  numbers  have  the  following  logarithms  ? 
17.  0.  18.  i.  19.  -  2.  20.  f .  21.  -  1. 

Principles  of  Logarithms. 

4.  The   logarithm  of  1    to   any   base   is   0.     For  6°  =  1,  or 
log,  1  =  0. 

5.  The   logarithm   of  the   base  itself  is   1.      For   bl  =  b,   or 
log*  6  =  1. 

6.  The  logarithm  of  a  product  is  equal  to  the  sum  of  the  loga- 
rithms of  its  factors  ;  or, 

log&  (m  x  ri)  =  Iog6  m  -f  log&  n. 


3-7]  LOGARITHMS.  375 

Let  Iog6  m  =  x  and  Iog6  n  =  y ; 

then      bx  =  m  and  6y  =  n,  and  therefore,  mn  =  bxby  =  6I+y. 
Translated  into  the  language  of  logarithms,  this  result  reads 

Iog6(rari)  =  x  +  y. 

But  x  =  Iog6  m  and  y  =  Iog6  ?i, 

and  consequently 

Iog6(mw)  =  Iog6  m  +  Iog6  n, 

for  all  positive  values  of  b. 

This  result  may  be  readily  extended  to  a  product  of  any 
number  of  factors.     For, 

logb(mnp)  =  logb(mn)  +  logsp  =  Iog6  m  +  Iog5  n  +  logbp. 
And,  in  like  manner,  for  any  number  of  factors. 

E.g.     Given   Iog2  32  =  5,  and   Iog2  64  —  6 ;  what  is  the  loga- 
rithm of  2048  to  the  base  2  ? 
Since  2048  =  32  .  64,  we  have 

Iog2  2048  =  Iog232  +  Iog2  64  =  5  +  6  =  11 

7.    The  logarithm  of  a  quotient  is  equal  to  the  logarithm  of  the 
dividend  minus  the  logarithm  of  the  divisor;  or, 

Iog6  (m  +  n)=  logb  m  -  Iog6  n. 

Let  Iog6 m  =  x  and  Iog6 n  =  y; 

then  bx  =  m  and  by  =  n,  and  therefore  m  -i-  n  =  bx  -5-  by  =  bx~y. 

In  the  language  of  logarithms  the  last  equation  is 

Iog6  (m-r-ri)=x  —  y=.  Iog6  m  —  Iog5  n, 
for  all  positive  values  of  b. 

E.g.     Given  Iog3  3  =  1  and  Iog3  21 87  =  7,  what  is  the  loga- 
rithm of  729  to  the  base  3  ? 

Since  729  =  ^-^, 

we  have       Iog3  729  =  Iog3  2187  -  Iog3  3  =  7  -  1  =  6. 


376  ALGEBRA.  [Cn.  XXVI 

8.  Both  m  and  n  may  be  products,  or  the  quotient  of  two 
numbers. 

E.g.,     loglo  |^-f  =  loglo  (4  x  5)  -  loglo  (9  x  8) 

«/  X  o 

=  Iog10  4  +  Iog10  5  -  Iog10  9  -  Iog10  8. 

9.  The  logarithm  of  the  reciprocal  of  any  number  is  the  oppo- 
site of  the  logarithm  of  the  number. 

For,  Iog5  -  =  Iog6  1  -  Iog6  n 

n 

=  —  Iog6  n,  since  Iog6 1  =  0. 
E.g.,  Iog2  4  =  2,  and  Iog2  J  =  -  2. 

10.  TJie  logarithm  of  any  power,  integral  or  fractional,  of  a 
number  is  equal  to  the  logarithm  of  the  number  multiplied  by  the 
exponent  of  the  power  ;  or 

log(m^)  =  p  logm. 

Let  Iog6  m  =  x,  then  bx  =  m. 

Raising  both  sides  of  the  last  equation  to  the  pfh  power,  we 
have  bpx  =  mp,  or  Iog6  (mp)  =  px  =p  Iog6  m. 

E.g.,  If  Iog5  25  =  2,  what  is  Iog5  (25)3  ? 

We  have         Iog5  (25)3  =  3  log,  25  =  3  x  2  =  6. 

11.  When  the  exponent  is  a  positive  fraction  whose  numera- 
tor is  1,  this  principle  may  be  conveniently  stated  thus : 

The  logarithm  of  a  root  of  a  number  is  the  logarithm  of  tlic 
number  divided  by  the  index  of  the  root. 

For,  log,  (m«)  =  -  log  m  =  l2iL^:. 

q  q 

E.g.,  If  Iog7  2401  =  4,  what  is  Iog7  V2401  ? 

We  have         Iog7  V2401  =  -  Iog7  2401  =  -.4  =  2. 


8-12]  LOGARITHMS.  377 


EXERCISES  II. 


Express  the  following  logarithms  in  terms  of  log  a,  log  6, 

log  c,  and  log  d  : 


Express  the  following  sums  of  logarithms  as  logarithms  of 
products  and  quotients. 

8.   log  a  -f  log  b  —  log  c.  9.    log  a  —  (log  b  -f  log  c). 

10.   31oga-^log(6  +  c).       11.   i  log  (1  -  a)  +  flog  (!  +  *)• 


12.   21og-  +  31og-«  13.    21oga-f  logfc-f-ilogc. 

b  a 

Given  Iog10  2  =  .30103,  Iog10  3  =  :47712,  Iog10  5  =  .69897, 
Iog10  7  =  .84510,  find  the  values  of  the  following  logarithms, 
to  the  base  10  : 

14.  log  50.  15.    log  6.  16.    log  8.  17.    log  9. 

18.  log  12.  19.    log  36.          20.    log  108.  21.    log-4J-. 

22.  log  2}.  23.    logSf.  24.    Iog5-f.  25.    log  360. 

26.  log  3072.  27.   log  3500.  28.  log  5880. 

29.  log  ^72.  30.    log  VI80-  31.  log  ^1715. 

32     1o^490  33     lo^^9^XVl05  j 

32.    log.  33.    logVV.          34.    log 


(4J)» 
, 

Systems  of  Logarithms. 

12.    The  two  most  important  systems  of  logarithms  are  : 
(i.)  The  system  whose  base  is  10.     This  system  was  intro- 

duced, in  1615,  by  the  Englishman,  Henry  Briggs. 

Logarithms  to  the  base  10  are  called  Common,  or  Briggs  's 

Logarithms. 


378  ALGEBRA.  [Cn.  XXVI 

(ii.)  The  system  whose  base   is   the  sum  of  the  following 
infinite  series, 


The  value  of  this  sum,  which  to  seven  places  of  decimals  is 
2.7182818,  is  denoted  by  the  letter  e. 

Logarithms  to  the  base  e  are  called  Natural  Logarithms; 
sometimes  also  Napierian  Logarithms,  in  honor  of  the  inventor 
of  logarithms,  the  Scotch  Baron  Napier,  a  contemporary  of 
Briggs.  Napier  himself  did  not,  however,  introduce  this  sys- 
tem of  logarithms. 

These  two  systems  are  the  only  ones  which  have  been  gen- 
erally adopted ;  the  common  system  is  used  in  practical  calcu- 
lations, the  natural  system  in  theoretical  investigations.  The 
reason  that  in  all  practical  calculations  the  common  system  of 
logarithms  is  superior  to  other  systems  is  because  its  base  10 
is  also  the  base  of  our  decimal  system  of  numeration. 

The  logarithms  of  most  numbers  are  irrational,  and  thus 
approximate  values  are  used. 

Properties  of  Common  Logarithms. 

13.   In   the  following  articles  the   subscript   denoting   the 
base  10  will  be  omitted. 
We  now  have 

10°=        1,  orlogl        -0; 

10'  =      10,  or  log  10      =  1; 

102  =    100,  or  log  100    =  2 ; 

103  =  1000,  or  log  1000  =  3  ; 


(a) 


10-1  =  .1,  or  log  .1  =  -  1 ; 
10-2'=  .01,  or  log  .01  =-2; 
10-3=  .001,  or  log .001  =-3; 
10  4  =  .0001,  or  log  .0001  =  -  4 ; 


12-15]  LOGARITHMS.  379 

Evidently  the  logarithms  of  all  positive  numbers,  except 
positive  and  negative  integral  powers  of  10,  consist  of  an  inte- 
gral and  a  decimal  part.  Thus,  since  101  <  85  <  102,  we  have 
1  <  log  85  <  2,  or  log  85  =  1  -f  a  decimal. 

14.  The  integral  part  of  a  logarithm  is  called  its  Character- 
istic. 

The  decimal  part  of  a  logarithm  is  called  its  Mantissa. 

15.  Since  a  number  having  one  digit  in  its  integral  part,  as 
7.3,  lies  between  10°  and  101,  it  follows  from  table  (a)  that  its 
logarithm  lies  between  0  and  1,  i.e.,  is  0  +  a  decimal.     Since  any 
number  having  two  digits   in  its  integral  part,  as  76.4,  lies 
between  101  and  102,  its  logarithm  lies  between  1  and  2,  that  is, 
is  1  +  a  decimal     In  general,  since  any  number  having  n  digits 
in  its  integral  part  lies  between  10""1  and  10n,  its  logarithm 
lies  between  n  —  1  and  n,  i.e.,  is  n  —  1  +  a  decimal.     We  there- 
fore have : 

(i.)  The  characteristic  of  the  logarithm  of  a  number  greater 
than  unity  is  positive,  and  is  one  less  than  the  number  of  digits  in 
its  integral  part. 

E.g.,  log  2756.3  =  3  +  a  decimal. 

Since  a  number  less  than  1  having  no  cipher  immediately 
following  the  decimal  point  lies  between  10°  and  lO"1,  it  follows 
from  table  (6)  that  its  logarithm  lies  between  0  and  —  1,  i.e., 
is  —  1  4-  a  positive  decimal.  Since  a  number  less  than  1  having 
one  cipher  immediately  following  the  decimal  point  lies  between 
10-1  and  10-2,  its  logarithm  lies  between  —  1  and  —  2,  i.e., 
is  —  2  -f  a  positive  decimal.  In  general,  since  a  number  less 
than  1  having  n  ciphers  immediately  following  the  decimal 
point  lies  between  W~n  and  10-(M+1),  its  logarithm  lies  between 
-  n  and  —  (n  -f  1),  i.e.,  is  —  (n  +  1)  +  a  positive  decimal.  We 
therefore  have : 

(ii.)  The  characteristic  of  the  logarithm  of  a  number  less  than 
1  is  negative,  and  is  numerically  one  greater  than  the  number  of 
ciphers  immediately  following  the  decimal  point. 

E.g.,  log  .00035  =  —  4  +  a  decimal  fraction. 


380  ALGEBRA.  [Cn.  XXVI 

It  follows  conversely  from  (i.)  and  (ii.) : 

(iii.)  If  the  characteristic  of  a  logarithm  be  -\-n,  there  are  n-f-1 
digits  in  the  integral  part  of  the  corresponding  number. 

(iv.)  If  the  characteristic  of  a  logarithm  be  — n,  there  are  n  —  1 
ciphers  immediately  following  the  decimal  point  of  the  correspond- 
ing number. 

16.  It  lias  been  found  that  538  =  102-73078  to  four  decimal 
places,  or  log  538  =  2.73078.     We  also  have 

log  .0538  =  log  ySf  for  =  log  538  -  log  10000  =  2.73078  -  4 

=  .73078  -  2 ; 
log  5.38  =  log  fff  =  log  538  -  log  100  =  2.73078  -  2 

=  .73078; 
log  53800  =  log  (538  x  100)  =  log  538  +  log  100 

=  2.73078  +  2  -  4.73078. 

These  examples  illustrate  the  following  principle : 

If  two  numbers  differ  only  in  the  position  of  their  decimal 

points,  their   logarithms   have   different  characteristics   but   the 

same  positive  mantissa. 

17.  The  characteristic  and  the  mantissa  of  a  number  less 
than  1  may  be  connected  by  the  decimal  point,  if  the  sign  (— ) 
be  written  over  the  characteristic  to  indicate  that  the  character- 
istic only  is  negative,  and  not  the  entire  number. 

Thus,  instead  of  log  .00709  =  .85065  -  3  =  -  3  +  .85065,  we 
may  write  3.85065 ;  this  must  be  distinguished  from  the  ex- 
pression —  3.85065,  in  which  the  integer  and  the  decimal  are 
both  negative.  Similarly, 

log  .082  =  2.91381,  while  log  .820  =  2.91381. 

Five-Place  Table  of  Logarithms. 

18.  The  logarithms,  to  the  base  10,  of  a  set  of  consecutive 
integers  have  been  computed. 

In  tabulating  these  logarithms,  compactness  is  important. 


15-20]  LOGARITHMS.  381 

For  this  reason,  all  unnecessary  detail  is  omitted.  Since  the 
characteristic  of  the  logarithm  of  any  number  can,  as  we  have 
seen,  be  determined  by  inspection,  it  is  unnecessary  to  write  it 
with  the  mantissa  in  the  table.  Consequently,  only  the  man- 
tissas, without  the  decimal  points,  are  there  given. 

Neither  is  it  necessary  to  give  the  logarithms  of  decimal 
fractions,  since  their  mantissas  are  the  same  as  the  mantissas 
of  the  numbers  obtained  by  omitting  the  decimal  point. 

The  logarithms  may  be  carried  to  any  number  of  decimal 
places,  and  the  extent  to  which  they  are  carried  depends  upon 
the  degree  of  accuracy  required  in  their  use. 

19.  The  accompanying  five-place  table  gives  the  mantissas 
of  the  logarithms  of  all  consecutive  integers  from  1  to  9999 
inclusive. 

In  this  table  the  first  three  figures  of  each  number  are 
given  in  the  column  headed  N,  and  the  fourth  figure  in  the 
horizontal  line  over  the  table.  The  first  figure,  which  is  the 
same  for  all  numbers  in  a  given  column,  is  printed  in  every 
tenth  number  only. 

The  columns  headed  0,  1,  2,  3,  etc.,  contain  the  mantissas, 
with  decimal  points  omitted. 

In  the  column  headed  0,  when  the  first  two  figures  are  not 
printed,  they  are  to  be  taken  from  the  last  mantissa  above 
which  is  printed  in  full. 

In  the  columns  headed  1,  2,  3,  etc.,  the  last  three  figures 
only  are  printed ;  the  first  two  are  to  be  taken  from  the  column 
headed  0  in  the  same  horizontal  line. 

When  a  star  is  prefixed  to  the  last  three  figures  of  a  man- 
tissa, the  first  two  figures  are  to  be  taken  from  the  line  below. 

To  Find  the  Logarithm  of  a  Given  Number. 

'  20.  When  the  Number  consists  of  Four  or  Fewer  Figures.  — 
Take  the  mantissa  that  is  in  the  horizontal  line  with  the  first 
three  figures  and  in  the  column  under  the  fourth  figure  of  the 
given  number. 

Determine  the  characteristic  by  Art.  15. 


382  ALGEBRA.  [Cn.  XXVI 

E.g.,    log   2583  =  3.41212,  log  46.32  =±  1.66577. 

In  writing  logarithms  with  negative  characteristics  it  is 
customary  to  modify  the  characteristics  so  that  10  is  uniformly 
subtracted  from  the  logarithms. 

Thus,        2.45926  =  .45926  -  2  =  8.45926  -  10 ; 
4.37062  =  .37062  -  4  =  6.37062  -  10. 

That  is,  we  add  10  to  the  negative  characteristic,  and  write 
—  10  after  the  logarithm. 

log  .5757  =  9.76020  -  10,  log  .02768  =  8.44217  -  10. 

Observe  that  the  first  two  figures  of  the  mantissa  of  log 
.5757  are  taken  from  the  line  below,  in  accordance  with  the 
directions  in  Art.  19. 

If  the  given  number  consists  of  fewer  than  four  figures, 
annex  ciphers  until  it  has  four  figures,  in  taking  the  mantissa 
from  the  table. 

E.g.,    mantissa  of  log  78  =  mantissa  of  log  7800  =  .89209, 
and  log  78  =  1.89209. 

In  like  manner, 

log  583  =  2.76567,  log  .02  =  8.30103  -  10. 

21.  When  the  Number  consists  of  more  than  Four  Significant 
Figures.  — The  method  used  is  called  interpolation,  and  depends 
upon  the  following  property  of  logarithms: 

The  difference  between  two  logarithms  is  very  nearly  propor- 
tional to  the  difference  between  the  corresponding  numbers  when 
this  difference  is  small. 

.  The  error   made   by   assuming   that  these   differences   are 
exactly  proportional  will  be  so  small  that  it  may  be  neglected. 

Ex.  1.     Find  log  27845. ' 

Omitting,  for  the  moment,  the  decimal  points  from  the 
mantissas,  we  have 

mantissa  of  log  27850  =  44483, 

mantissa  of  log  27840  =  44467, 

difference  of  mantissas  =        16. 


20-21]  LOGARITHMS.  383 

Let  x  stand  for  the  difference  between  the  mantissas  of  log 
27845  and  log  27840 ;  that  is,  for  the  correction  to  be  added  to 
the  smaller  mantissa  to  give  the  required  mantissa. 

Then,  by  the  above  property, 

x  =  27845  -  27840  =  5        - 
16     27850  -  27840  ~~  10  " 
Whence  x  =  .5  x  16  =  8. 

Therefore,  mantissa  of  log  27845  =  44467  +  8  =  44475, 
and  log  27845  =  4.44475. 

Observe  that,  by  Art.  16,  the  mantissa  of  log  27850  is  the 
same  as  the  mantissa  of  log  2785.  In  subsequent  work  such 
ciphers  will  be  omitted. 

The  method  can  now  be  stated  more  concisely  for  practical 
work : 

/Subtract  the  mantissa  corresponding  to  the  first  four  figures  of 
the  given  number  from  the  next  mantissa  in  the  table  ;  multiply 
this  difference  by  the  remaining  figure  or  figures  of  the  given 
number,  treated  as  a  decimal  ;  add  the  product  to  the  first  (and 
smaller)  mantissa. 

Prefix  finally  the  proper  characteristic. 

In  thus  finding  the  mantissa,  a  decimal  point  in  the  given 
number  is  ignored,  in  accordance  with  Art.  16. 

The  difference  between  two  consecutive  mantissas  in  the  table 
is  called  the  Tabular  Difference. 

Ex.  2.   Find  log  78.1283. 
We  have    mantissa  of  log  7813  =  89282, 
mantissa  of  log  7812  =  89276, 
tabular  difference  =          6, 

correction  ==  .83  x  6  =  4.98, 
mantissa  of  log  781283  =  89276  +  5  =  89281. 
Therefore  log  78.1283  =  1.89281 

Observe  that  the  correction  added  to  the  mantissa  of  log 
7812  is  5,  the  nearest  integer  to  4.98. 


384  ALGEBRA.  [Cn.  XXVI 

22.  In  the  table  of  logarithms  a  column  containing  the 
required  corrections  (head  Pp.  Pts.,  i.e.,  proportional  parts)  is 
given.  In  this  column  there  are  several  small  tables,  each  con- 
taining two  columns  of  numbers.  One  of  these  columns  consists 
of  the  consecutive  numbers  1  to  9 ;  the  other,  headed  by  a  tabu- 
lar difference,  contains  the  correction  corresponding  to  each  one 
of  the  figures  1  to  9,  when  it  is  the  fifth  figure  of  the  number 
whose  logarithm  is  required.  When  it  is  the  sixth  figure,  the 
corresponding  tabular  correction  must  evidently  be  divided  by 
10 ;  when  it  is  the  seventh  figure,  by  100 ;  and  so  on. 

Thus,  in  Ex.  1  of  the  preceding  article,  we  take  the  correc- 
tion opposite  5,  under  the  tabular  difference  16,  and  obtain  8, 
as  before. 

In  Ex.  2,  we  take  the  following  corrections  from  the  column 
headed  by  the  tabular  difference  6 : 

for  8,  correction  =  4.8 

for  3,  correction  =  0.18 

final  correction  =  4.98,  as  before. 

Observe  that  the  correction  for  the  sixth  figure  of  the  given 
number  does  not  affect  the  result. 

Ex.  3.    Find  the  log  .0128546. 
We  have         mantissa  of  log  1286  =  10924, 
mantissa  of  log  1285  =  10890, 
tabular  difference  =        34. 

From  the  column  of  proportional  parts  headed  by  34,  we 
obtain : 

correction  for  fifth  figure  4  =  13.6 

correction  for  sixth  figure  6  =    2.04 
total  correction  =  15.64 

Therefore,  mantissa  of  log  128546  =  10890  +  16  =  10906, 
and  log  .0128546  =  8.1090.6  -  10. 

Observe  that  in  this  example  the  correction  for  the  sixth 
figure  does  affect  the  result. 


22-24]  LOGARITHMS.  385 

EXERCISES   III. 

Verify  the  following  statements : 
1.   log  13  =  1.11394.  2.   log  14.84  =  1.17143. 

3.   log  73000  =  4.86332.  4.   log  5884.4  =  3.76970 

5.  log  .031586  =  8.49949  -  10. 

6.  log  .00391857  =  7.59S13  -  10. 

Find  the  logarithms  of  each  of  the  following  numbers : 
7.   5.  8.   18.  9.   540.  10.   3876. 

11.   2076.  12.   59.80.          13.   1.87.          14.    .01832. 

15.   .0004129.      16.   63072.         17.   59.836.      18.    4376.4. 
19.    .070518.  20.  185462.  21.    .00103987. 

To  find  a  Number  from  its  Logarithm. 

23.  Mantissa  given  in  the  Table.  —  If  the  mantissa  of  the 
given  logarithm  is  found  in  the  table,  the  first  three  figures 
of  the  required  number  will  be  in  the  same  line  with  it  in  the 
column  headed  N,  and  the  fourth  figure  over  the  column  in 
which  the  given  mantissa  stands. 

The  characteristic  is  determined  by  Art.  15  (iii.)  and  (iv.). 

Ex.  1.  Find  the  number  whose  logarithm  is  4.82099.  The 
mantissa  .82099  corresponds  to  the  number  6622;  but  since 
the  given  characteristic  is  4,  the  required  number  must  have 
five  integral  places. 

Consequently  4.82099  =  log  66220. 

Ex.  2.  Find  the  number  whose  logarithm  is  8.78625  — 10. 
The  mantissa  .78625  corresponds  to  the  number  6113;  but 
since  the  characteristic  is  —  2,  the  required  number  must  be 
a  decimal  having  its  first  significant  figure  in  the  second  deci- 
mal place. 

Consequently          8.78625  -  10  =  log  .06113. 

24.  Mantissa  not  given  in  the  Table.  —  The  method  employed 
is  the  converse  of  that  used  in  Art.  21  to  find  the  logarithms  of 
numbers  that  consist  of  more  than  four  significant  figures. 


386  ALGEBRA.  [Cn.  XXVI 

Ex.  1.   Find  the  number  whose  logarithm  is  2.81727. 
We  have 

given  mantissa  =  81727 ; 

next  smaller  mantissa  =  81723,  corresponding  number  =  6565 ; 
next  larger  mantissa  =  81730,  corresponding  number  =  6566. 

Let  x  stand  for  the  difference  between  6565  and  the  required 
number ;  that  is,  for  the  correction  to  be  added  to  6565. 
We  then  have 

x  =  81727  -  81723    Qr  x  =  ±=  6 

6566  -  6565  '•    81730  -  81723'  '  T  1 "  7  " 

corrected  for  the  first  decimal  place.  Notice  that  the  signifi- 
cance of  the  decimal  point  in  the  result  is  that  the  correction 
is  to  be  annexed  as  an  additional  figure  to  the  smaller  number. 

Therefore,  the  figures  in  the  required  number  are  65656 ;  and 
since  the  characteristic  of  the  given  logarithm  is  2,  there  are 
only  three  integral  places.  Hence  2.81727  =  log  656.56. 

This  process  may  also  be  stated  concisely  for  practical  work  : 

Take  the  mantissa  next  smaller  and  the  mantissa  next  larger 
than  the  given  mantissa,  and  note  the  numbers  corresponding; 
next  divide  the  difference  between  the  given  mantissa  and  the  next 
smaller  by  the  difference  between  the  next  larger  and  the  next 
smaller.  Annex  the  quotient  to  the  number  corresponding  to  the 
smaller  mantissa,  neglecting  the  decimal  point  of  the  quotient. 

Place  the  decimal  point  in  the  number  thus  obtained  as  it  is 
determined  by  the  given  characteristic. 

Ex.  2.   Find  the  number  whose  logarithm  is  7.18281  —  10. 
We  have 

given  mantissa  =  18281 ; 

next  smaller  mantissa  =  18270,  corresponding  number  =  1523 ; 
next  larger  mantissa  =  18298,  corresponding  number  =  1524. 
Hence  the  correction  to  be  annexed  to  1523  is 

18281  -  18270     =  11     =  39  , 
18298-18270'       28' 


24-25]  LOGARITHMS.  387 

Therefore  the  figures  of  the  required  number  are  152339 ; 
and  since  the  characteristic  of  the  given  logarithm  is  —  3, 
there  must  be  two  ciphers  between  the  decimal  point  and  the 
first  significant  figure. 

Consequently    7.18281  -  10  =  log  .00152339. 

In  general,  in  using  a  five-place  table,  the  numbers  corre- 
sponding to  given  mantissas  should  be  carried  to  only  Jive 
significant  figures,  as  in  Ex.  1. 

But  with  mantissas  in  the  first  two  pages  of  the  table,  the 
corresponding  numbers  may  be  carried  to  six  figures.  The 
reason  being  that  the  tabular  differences  later  become  so  small 
that  the  correction  for  a  sixth  figure  will  not  in  general  affect 
the  result.  See  Exx.  2-^3,  Art.  22. 

25.  The  correction  to  be  added  to  the  number  corresponding 
to  the  next  smaller  mantissa  may  also  be  taken  from  the 
column  of  proportional  parts. 

In  this  column  turn  to  the  table  headed  by  the  number 
which  is  equal  to  the  difference  between  the  next  larger  and 
the  next  smaller  mantissa.  As  the  first  figure  of  the  correc- 
tion take  the  figure  in  this  table  which  is  opposite  the  pro- 
portional part  nearest  to  the  difference  between  the  given 
mantissa  and  the  next  smaller  mantissa. 

If  a  second  figure  in  the  correction  is  to  be  found,  we  should 
take  as  the  first  figure  that  figure  which  is  opposite  the  pro- 
portional part  next  smaller  than  the  difference  between  the 
given  mantissa  and  the  next  smaller. 

Multiply  by  10  the  difference  between  the  proportional  part 
already  used  and  the  difference  between  the  given  mantissa 
and  the  next  smaller,  and  take  the  product  as  a  proportional 
part  in  determining  the  second  figure  of  the  correction ;  and 
so  on. 

Thus,  in  Ex.  1  of  the  preceding  article,  we  turn  to  the 
column  headed  by  the  tabular  difference  7.  The  proportional 
part  in  this  table  that  is  nearest  to  4  (the  difference  between 
the  given  mantissa  and  the  next  smaller)  is  4.2 ;  the  number 
opposite  4.2  is  6,  the  correction  previously  obtained. 


388  ALGEBRA.  [Cn.  XXVI 

In  Ex.  2,  we  turn  to  the  column  headed  by  the  tabular  differ- 
ence 28.  The  proportional  part  next  smaller  than  11  (the 
difference  between  the  given  mantissa  and  the  next  smaller) 
is  8.4;  the  figure  opposite  8.4  is  3,  the  first  figure  of  the 
correction. 

We  next  multiply  2.6  (=11  —  8.4)  by  10,  and  take  the 
product  26  as  a  proportional  part.  The  figure  opposite  25.2 
(nearest  to  26)  in  the  column  headed  by  28  is  9,  the  second 
figure  of  the  correction.  Therefore,  the  required  correction  is 
found  to  be  39,  as  before. 

EXERCISES   IV. 

Verify  the  following  statements : 

1.  log  x  =  3.14926,  x  =  1410.13. 

2.  log  x  =  1.59187,  x  =  39.073. 

3.  log  x  =  . 34159,  x  =  2.1958. 

4.  log  x  =  9.57187  -  10,  x  =  .37314. 

5.  log  x  =  7.83957  -10,  x  =  .0069115. 
€.  log  x  =  6.18953  -  10,  x  =  .00015471. 

Find  the  numbers  whose  logarithms  are : 

7.   2.26150.                  8.   .59726.  9.  8.94655-10. 

10.   3.88825.                11.   6.19815.  12.  6.72576-10. 

13.   4.98880.                14.   1.68417.  15.  9.23360-10. 

Cologarithms. 

26.  The  Cologarithm  of  a  number,  or,  as  it  is  sometimes 
called,  the  Arithmetical  Complement  of  the  logarithm,  is  de- 
fined as  the  logarithm  of  the  reciprocal  of  the  number. 

That  is,  colog  n  —  log  -  =  log  1  —  log  n  =  0  —  log  n. 
n 

We  thus  see  that  the  cologarithm  of  a  number  is  obtained 
by  subtracting  its  logarithm  from  0.  But  this  step  would 
leave  the  mantissa  as  well  as  the  characteristic  negative.  To 
avoid  a  negative  mantissa,  therefore,  we  subtract  the  logarithm 
from  10  -  10,  =  0. 


25-27]  LOGARITHMS.  389 

Ex.  1.    Find  the  colog  3. 

Subtracting  log  3,  =  .47712,  from  10  —  10,  we  have 
10.  - 10 

.47712 

9.62288  -  10 

Therefore  colog  3  =  9.62288  -  10. 
Ex.  2.   Find  colog  .0054. 

Subtracting  log  .0054,  =  7.73239  -  10,  from  10  -  10,  we  have 
10.  - 10 

7.73239  - 10 
2.26761 
Therefore  colog  .0054  =  2.26761. 

EXERCISES  V. 

Verify  the  following  statements  : 

1.  colog  543          =7.26520-10. 

2.  colog  72.318      =8.14075-10. 

3.  colog  8.9134      =9.04996-10. 

4.  colog  .38145      =.41856. 

5.  colog  .051984    =1.28413. 

6.  colog  .0091437  =  2.03887. 

Find  the  cologarithm  of  each  of  the  following  numbers : 

7.   5817.  8.   .6305.  9.    .009812.       10.   763.85. 

11.    15.482.        12.   7.00386.         13.   .000594.      14.   32581.9 

Applications. 

27.   Ex.  1.  Compute  the  value  of  a?,  when 
x  =  53.847  x  .0085965.    . 
log  x  =  log  53.847  +  log  .0085965. 

log  53.847  =  1.73117 
log  .0085965=  7.93433  -  10 
log  x  =  9.66550  —  10 

x  =  .46291. 


390  ALGEBRA.  [Cn.  XXVI 

Ex.  2.   Compute  the  value  of  x,  when 
x  =  8.4394  --  .31416. 
log  x  =  log  8.4394  +  colog  .31416. 

log  8.4394=    .92631 

colog .31416  =    .50285 

log  i x  =  1.42916 

x  =  26.863. 

Ex.  3.    Compute  the  value  of  a?,  when 

=  6.4319  x  .59218 

7.9254  x  .062547* 

log  x  =  log  6.4319  +  log  .59218  +  colog  7.9254  +  colog  .062547. 
log  6.4319  =      .80834 
log  .59218=    9.77246-10 
colog  7.9254  =    9.10098  -  10 
colog  .062547  =    1.20379 

log  x  =  20.88557  -20 

=      .88557. 
x  =  7.6837. 

Ex.  4.    Find  the  value  of  a?,  when 

x  =  .53184. 
log  x  =  4  log  .5318 

=  4(9.72575-10) 
=  38.90300  -  40 
=    8.90300  -  10. 
x  =  .079983. 

Ex.5.    Find  the  value  of  ty-.  031459. 

Since  a  negative  number  cannot  be  expressed  as  a  power  of 
+ 10,  such  a  number  does  not  have  a  logarithm.  In  this 
example,  therefore,  and  in  all  similar  examples,  we  first  deter- 
mine the  sign  of  the  result.  We  then  find  the  value  of  the 
expression  obtained  by  changing  each  sign  —  to  +,  and  to  that 
result  prefix  the  sign  previously  determined. 


27]  LOGARITHMS.  391 

The  sign  of  the  result  of  this  example  is  — 
Let  a;  =  y. 031459. 

Then  log  x  =  1  log  .031459 

=  1(28.49775-30) 

=  9.49925  - 10, 
and  x  =  .31568. 

Therefore,  the  required  result  is  —  .31568. 

Observe  that  in  dividing  log  .031459  by  3,  we  first  modified 
the  characteristic  so  that  the  number,  30,  which  is  subtracted 
from  the  logarithm  is  10  times  the  divisor ;  that  is,  so  that  the 
quotient  obtained  by  dividing  this  number  by  3  is  10. 

Ex.  6.   Compute  the  value  of  x,  when 

=  4.5921  x  -C/.021946 
=  .059318  x  .415873  ' 

log  x  =  log  4.5921  + 1  log  .021946  +  colog  .059318 

+  3  colog  .41587. 

log  4.5921  =      .66201 

i  log  .021946  =  i  (28.34135  -  30)  =  9.44712  -  10 

colog  .059318  =  1.22681 

3  colog  .41587  =  3  x  .38104  =  1.14312 

log  x  =  12.47906  -10 

=  2.47906. 
x  =  301.34. 

Ex.  7.   Compute  the  value  of  x,  when 

3/5.4318  x  y.31459 


7.1938  x  .29342 

For  convenience  in  arranging  the  logarithmic  work,  we  first 
cube  both  members  of  this  equation,  and  obtain 

^  =  5.4318  x  y.31459  m 

7.1938  x  .2934s 


392  ALGEBRA.  [Cn.  XXVI 

Taking  logarithms,  we  have 

3  log  x  =  log  5.4318  +  i  log  .31459  +  colog  7.1938 

+  2  colog  .2934.      (2) 

In  practice,  step  (1)  should  be  performed  mentally,  and  the 
result  (2)  be  at  once  written. 

log  5.4318=  .73494 

i  log  .31459  =  1(19.49775  -  20)  =  9.74887  -  10 

colog  7.1938  =  9.14304  -  10 

2  colog  .2934  =  2  x  0.53254  =  1.06508 

3  log  x  =  20.69193  -20 

=  .69193. 
log  x=.  23064. 
x  =  1.70076. 

EXERCISES   VI. 

Find  the  values  of  each  of  the  following  expressions  : 
1.   31.834  x  185.592.  2.   8.0043  x  .5319. 

3.   .004893  x  6.5942.  4.    (-  .0514)  x  .123857. 


5     '78.                              6 

1539 

19.7939 

'   347 
380.14  x  (-  .0576) 

78395 
9. 

11 

3892.7 
(-  9.7408)  x  .000395 

7.3792 
5.83  x  91.358 

36.937 
57.13  x  9.0047 

.00479  5.382  x  .07235 

12    4.9  x  (-306)  x  48.3  .79 x 891.3 x. 00099 

100.088  x  2.9  x  .081 '  '  (-10.236)  x  .07  x  .0031 

14.    7.04353.  15.   .318444.  16.    2.38175. 

17.    (3.68  x  .97)4.  18.  (.7918  x  3.17)5. 

19.    [.034  x  (-  4.973S)]4.  20.  (17.19  x  .00001986)5. 


27*28]  LOGARITHMS.  393 

21.    ^/13.  22.    J/-251.  23.    -^39.837, 

24.     ^163.43.  25.    sX-314922;  26.     -«/1.0031. 

27.     ^/||.  28.    -4/11  29« 

30.    l^/f.  31.    21^/1  32. 

33.    (.74sy8.21)4.  34.    (5.21-^/.3817)6. 

35.   f-^'-  5x^/17.  36.   3f  ^/.38  x  .{/7.3815. 

37.    ^(-25V3)-  38-    ^/(H2.34-^/.003914). 

39.    ^/(17.2^/.718).  40.    ty(-  23^7.18943). 

41.   5.341^/(27.39^/.1439).     42.   23.4912-^(.l  8-^/17.3). 

c  /3.1 
\   .5 


-  9.2614 
.519VH7.38 


Exponential  Equations. 

28.  An  Exponential  Equation  is  an  equation  in  which  the 
unknown  number  appears  as  an  exponent  of  a  known  or  an 
unknown  number,  as  a*  =  b. 

Ex.  1.     Solve  the  equation     3X  =  9. 
Taking  logarithms,  x  log  3  =  log  9  =  2  log  3. 
Hence  x  =  2. 

This  result  could  have  been  obtained  by  inspection,  by 
writing  the  given  equation  3X  =  32. 

Ex.  2.    Find  the  value  of  x  in  3*  =  5. 


taking  logarithms,          x  log  3  =  log  5  ; 

log  5 
T~  ~ 
log  3  .47712 


log  5      .69897 
whence  X  =  ~  ~  =  —     ~  =  1.46497. 


394  ALGEBRA.  [Cn.  XXVI 

Ex.  3.   Find  the  value  of  x  in  the  following  equation 


taking  logarithms,  (3  x  +  1)  log  2  =  (2  x  -  1)  log  7. 

Removing  parenthesis,  3  x  log  2  +  log  2  =  2  x  log  7  —  log  7, 
or  o?(3  log  2  -  2  log  7)  =  -  log  7  -  log  2; 

whence  g        log  7  +  log  2 

2  log  7  -  3  log  2 

.84510  4-  .30103 


1.69020  -  .90309 
1.14613 


.78711 


=  1.4561. 


EXERCISES  VII. 

Solve  the  following  exponential  equations : 

1.  2*  =  64.  2.   3'  =  SI.                3.   2*~l  =  .&*-*. 

4.  (_8)-*=16.  5.   48-1  =  .5-8.          6.   4*  =  8. 

7.  8X=32.  8.   5X  =  (V5)-1.         9.   4I+1  =  8-2'+2. 

10.  253*-1  =  625  •  5*+3.  11.   7^(I-3)  =  343-1  •  49v"*-3>. 

12.  27^(I-3)  =  ( V3)V(x+8).  13.    Vall~x  =  a8~z- 

14.  ^/a^2-VaX~3-  15-    V«3~4j:^^«<5~7a:xal=l. 

16.  (i)z=25.  17.    (i)-7=64. 

18.  (fj)li;!-fi=(^)78-8.  19-    (|)4*-7  =  .752-3*. 

20.  4^-6.2^  +  8  =  0.  21.   9*  +  243  =  36  •  3*. 

22.  3'°sz  =  9.            23.  5log2*  =  625.            24.    16log3x  =  32IO«x. 

25  ;V  =  10.  26.    16*  =  45.        27.    11"  =  310. 

28.  25*  =10.  29.    7*  =  300.      30.   3.594*  =  359600. 

31.  ^/9.8926  =  1.29.  32.   5*  =  73-14.       33.    aV2  =  ^/3. 

34.  r>*+3=1000.  35.   7X+1=5.       36.    1.58*-5=  9.847. 

37.    5*+1  =  II-1.  38.    3*+7=7*+3. 

39.   31*+3  =  2tV+4.  40.   35qe+2  =  40— *. 


28-32]  LOGARITHMS.  395 

Compound  Interest  and  Annuities. 

29.  To  find  the  compound  interest,  /,  and  the  amount,  A,  of 
a  given  principal,  P,  for  n  years  at  r  per  cent. 

If  the  interest  is  payable  annually,  the  amount  of  $  1  at  the 
end  of  one  year  will  be  1  -f  r  dollars,  and  the  amount  of  P 
dollars  will  be  P(l  +  r)  dollars.  This  amount,  P(l  +  r),  be- 
comes the  principal  at  the  beginning  the  second  year.  There- 
fore, at  the  end  of  the  second  year  the  amount  will  be 
P(l  +  r)  X  (1  4-  ?•),  =  P(l  4-  rf  dollars,  and  so  on. 

Therefore,  at  the  end  of  n  years  the  amount  will  be  P(l  4-  ?*)" 
dollars,  or 

*=/»(! +/•)". 

30.  This  formula  can  be  used  not  only  to  find  A,  but  also 
to  find  P,  r,  or  n,  when  the  three  other  quantities  are  given. 
Thus, 


P  = 


(i  + 


31.  An  Annuity  is  a  fixed  sum  of  money,  payable  yearly,  or 
at  other  fixed  intervals,  as  half-yearly,  once  in  two  years,  etc. 

32.  To  find  the  present  value,  P,  of  an  annuity  of  A  dollars, 
payable  yearly  for  n  years,  at  r  per  cent. 

The  present  *vorth  of  the  first  payment  is  dollars,  of 

A  IT? 

the  second  payment  is  — dollars,  and,  in  general,  of  the 

nth  payment  is dollars. 

(l  +  r)n 

Therefore  the  present  worth  of  all  the  payments  is 
A 


+  „    ,  .**  + 


Multiplying  numerator  and  denominator  by  1  4-  r,  we  have 
~r|_'       (1  +  /•)"_] 


396  ALGEBRA.  [Cn.  XXVI 

Ex.  1.   Find  the  amount  of  $500  for  8  years  at  5%  com- 
pound interest. 

A  =  P  (I  +  r)n  =  500  x  1.058. 

log  A  =  log  500  +  8  log  1.05. 

log  500  =  2.69897 
8  log  1.05  =    .16952 

log  A  =  2.86849 
A  =  738.73. 
Therefore  the  require'd  amount  is  $  738.73. 

Ex.  2.   Find  the  present  value  of  an  annuity  of  $  1000  for 
6  years,  if  the  current  rate  of  interest  is  5%. 


(l-fr) 
We  will  first  compute  1.066, 

log(1.05)6  =  6.  log  1.05 
=  6  x  .02119 
=  .12714. 
(1.05)6  =  1.34012. 


We  then  have 


20000  X"84012 
X 


.05  1.34012 

log  P  =  log  20000  +  log  .34012  +  colog  1.34012. 

log  20000=    4.30103 
log  .34012=    9.53163-10 
colog  1.34012  =    9.87286  -  10 

log  P=  23.70552  -20 

=    3.70552. 
P=5076. 
Therefore  the  present  value  of  the  annuity  is  $  5076. 


32]  LOGARITHMS.  397 

EXERCISES   VIII. 
Find  the  amount  at  compound  interest : 

1.  Of  $3600  for  5  years  at  41%. 

2.  Of  $1875.50  for  8  years  at  5%. 

3.  Of  $  12,350  for  6  years  at  3J%. 

4.  Of  $21,580  for  7  years  4  months  at  4%. 

Find  the  principal  that  will  amount  to : 

5.  $7913  in  5  years  at  5%  compound  interest. 

6.  $14,770  in  10  years  at  k\%  compound  interest. 

7.  $11,290  in  8  years  at  4%  compound  interest. 

8.  $11,090  in  6  years  6  months  at  3%  compound  intere'st. 

9.  In  what  time,  at  4%,  will  $8010  amount  to  $11,400  at 
compound  interest  ? 

10.  In  what  time,  at  4J%,  will  $3530  amount  to  $5987,  if 
the  interest  is  compounded  semi-annually  ? 

Find  the  rate  of  compound  interest : 

11.  If  $  lllO.amounts  to  $  1640  in  8  years. 

12.  If  $  3750  amounts  to  $6070  in  14  years. 

Find  the  present  value  of  an  annuity : 

13.  Of  $  1000  for  10  years,  if  the  current  rate  of  interest 

is  4%. 

14.  Of  $  1250  for  8  years,  if  the  current  rate  of  interest  is 

41%. 

15.  Of  $  2500  for  10  years,  if  the  current  rate  of  interest 
is  5%. 

16.  Of  $3000  for  12  years,  if  the  current  rate  of  interest 
is  6%. 


100-149 


N. 

0 

1 

2  I  3 

4 

5 

6 

7 

8 

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Pp.  Pts. 

100 

oo  ooo 

043 

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217 

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326 

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703 

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787 

828 

870 

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953 

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02  119 

1  60 

202 

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284 

325 

366 

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490 

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22.O 

21.5 

21.0 

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612 

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735 

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383 

423 

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5°3 

543 

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34-4 

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743 

782 

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862 

902 

941 

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39-6 

38.7 

37-8 

110 

04  139 

179 

218 

258 

297 

336 

376 

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454 

493 

41 

40 

39 

ii 

532 

571 

610 

650 

689 

727 

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538 

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729 

767 

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108 

145 

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221 

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19-5 

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350 

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636 

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814 

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23 

991 

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11.4 

n.  i 

10.8 

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377 

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552 

587 

621 

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4 

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25 

691 

726 

760 

795 

830 

864 

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934 

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5 

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106 

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394 

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516 

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534 

564 

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3 

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495 

524 

554 

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6 

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761 

791 

820 

850 

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231 

260 

289 

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319 

348 

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493 

522 

551 

580 

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28.8 

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0 

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5 

6 

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Pp.  Pts. 

11 


150-199 


N. 

0 

1 

2 

3 

4 

5   6 

7 

8 

9 

Pp.  Pts. 

150 

17609 

638 

667  696 

725 

754 

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811 

840 

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51 

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213 

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8 

23.2 

22.4 

59 

20  140 

167 

194 

222 

249 

276 

3°3 

330 

358 

385 

9 

26.1 

25.2 

160 

412 

439 

466 

493 

520 

548 

575 

602 

629 

656 

27   o.fi 

61 

683 

710 

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763 

790 

817 

844 

871 

898 

925 

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2-  1 

2.6 

62 

952 

978 

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2 

63 

21  219 

245 

272 

299 

325 

352 

378 

405 

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484 

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537 

564 

590 

617 

643 

669 

696 

722 

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4 

10.8 

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10.4 

65 

748 

775 

801 

827 

854 

880 

906 

932 

958 

985 

5 

13.5 

13.0 

66 

22  Oil 

037 

063 

089 

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141 

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194 

220 

246 

6 

16.2 

15.6 

67 

272 

298 

324 

350 

376 

401 

427 

453 

479 

505 

7 

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18.2 

68 

531 

557 

583 

608 

634 

660 

686 

712 

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69 

789 

814 

840 

866 

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917 

943 

968 

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9 

24-3 

23-4 

170 

23045 

070 

096 

121 

147 

172 

198 

223 

249 

274 

71 

300 

325 

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376 

401 

426 

452 

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553 

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603 

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654 

679 

704 

729 

754 

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2 

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73 

803 

830 

855 

880 

905 

930 

955 

980 

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74 

24055 

080 

105 

130 

155 

1  80 

204 

229 

254 

279 

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4 

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IO.O 

75 

304 

329 

353 

378 

403 

428 

452 

477 

5°2 

527 

5 

12.5 

76 

551 

576 

601 

625 

650 

674 

699 

724 

748 

773 

6 

15.0 

77 

797 

822 

846 

87I 

895 

920 

944 

969 

993 

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7 

17-5 

78 

25042 

066 

091 

115 

164 

188 

212 

237 

261 

8 

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79 

285 

310 

334 

358 

382 

406 

43  i 

455 

479 

503 

9 

22-5 

180 

527 

551 

575 

600 

624 

648 

672 

696 

720 

744 

24 

23 

81 

768 

792 

816 

840 

864 

888 

912 

935 

959 

983 

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2-4 

2.3 

82 

26  007 

031 

055 

079 

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126 

150 

174 

198 

221 

2 

4.8 

4.6 

83 

245 

269 

293 

316 

340 

364 

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411 

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6.9 

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529 

553 

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600 

623 

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670 

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85 

717 

741 

764 

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8  if 

834 

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881 

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11.5 

86 

95  1 

975 

998 

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6 

14.4 

13.8 

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27  184 

207 

231 

254 

277 

300 

323 

346 

37° 

393 

7 

1  6.8 

16.1 

88 

416 

439 

462 

485 

508 

554 

577 

600 

623 

8 

19.2 

18.4 

89 

646 

669 

692 

715 

738 

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784 

807 

830 

852 

9 

21.6 

20.7 

190 

875 

898 

921 

944 

967 

989 

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22 

21 

91 

28  103 

126 

149 

171 

194 

217 

240 

262 

285 

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2.2 

2.1 

92 

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353 

375 

398 

421 

443 

466 

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533 

2 

4-4 

4.2 

93 

556 

578 

60  1 

623 

646 

668 

69I 

713 

735 

758 

6.6 

6.3 

94 

780 

803 

825 

847 

870 

892 

914 

937 

959 

9,8i 

4 

8.8 

8.4 

95 

29003 

026 

048 

070 

092 

115 

137 

159 

181 

203 

5 

II.O 

10.5 

96 

226 

248 

270 

292 

3H 

336 

358 

380 

403 

425 

6 

13.2 

12.6 

97 

447 

469 

491 

535 

557 

579 

60  1 

623 

645 

7 

15-4 

14.7 

98 

667 

688 

710 

732 

754 

776 

798 

820 

842 

863 

8 

17.6 

1  6.8 

99 

885 

907 

929 

973 

994 

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9  19.8  18.9 

N. 

O 

1 

2 

3 

4 

5 

6 

7  |  8  |  9 

Pp.  Pts. 

200-249 


in 


N. 

O 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Pp.  Pts. 

200 

30103 

125 

146 

1  68 

190 

211 

233 

255 

276 

298 

22 

21 

01 

320 

363 

384 

406 

428 

449 

471 

492 

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I   2.2 

2.1 

02 

535 

557 

578 

600 

621 

643, 

664 

685 

707 

728 

2   4*4 

4-2 

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75° 

771 

792 

814 

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878 

899 

920 

942 

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6.3 

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963 

984 

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4  8.8 

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8.4 

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31  175 

197 

218 

239 

260 

281 

302 

323 

345 

366 

5  1  1.0 

10.5 

06 

387 

408 

429 

45° 

471 

492 

513 

534 

555 

576 

6  13.2 

12.6 

07 

597 

618 

639 

660 

68  1 

702 

723 

744 

765 

785 

7  :5'4 

14.7 

08 

806 

827 

848 

869 

890 

911 

93  i 

952 

973 

994 

8  17.6 

16.8 

09 

32015 

035 

056 

077 

098 

118 

139 

1  60 

181 

201 

9  19.8 

18.9 

210 

222 

243 

263 

284 

305 

325 

346 

366 

387 

408 

20 

ii 

428 

449 

469 

490 

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572 

593 

613 

i 

2.O 

12 

634 

654 

675 

695 

715 

736 

756 

777 

797 

818 

2 

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13 

838 

858 

879 

899 

919 

940 

960 

980 

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6.0 

14 

33041 

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082 

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132 

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163 

183 

203 

224 

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4 

8.0 

15 

244 

264 

284 

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325 

341 

365 

385 

405 

425 

5 

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16 

445 

465 

486 

506 

526 

546 

566 

586 

606 

626 

6 

12.0 

17 

646 

666 

686 

706 

726 

746 

766 

786 

806 

826 

7 

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18 

846 

866 

885 

905 

925 

945 

965 

985 

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8 

16.0 

19 

34°44 

064 

084 

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124 

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163 

183 

203 

223 

9 

1  8.0 

220 

242 

262 

282 

301 

321 

34i 

361 

380 

400 

420 

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21 

439 

459 

479 

498 

518 

537 

557 

577 

596 

616 

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1.9 

22 

635 

655 

674 

694 

7*3 

733 

753 

772 

792 

811 

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23 

830 

850 

869 

889 

908 

928 

947 

967 

986 

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c.7 

24 

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064 

083 

102 

122 

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1  60 

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218 

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257 

276 

295 

315 

334 

353 

372 

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5 

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26 

411 

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449 

468 

488 

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526 

545 

564 

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27 

603 

622 

641 

660 

679 

698 

717 

736 

755 

774 

7 

13 

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28 

793 

813 

832 

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870 

889 

908 

927 

946 

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15 

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29 

984 

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17 

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230 

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192 

211 

229 

248 

267 

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342 

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380 

399 

418 

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455 

474 

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511 

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549 

568 

586 

605 

624 

642 

661 

680 

698 

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2 

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736 

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773 

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810 

829 

847 

866 

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922 

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959 

977 

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37  I07 

125 

144 

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181 

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218 

236 

254 

273 

5 

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36 

291 

310 

328 

346 

365 

383 

401 

420 

438 

457 

6 

10.8 

37 

475 

493 

511 

530 

548 

566 

585 

603 

621 

639 

7 

12.6 

38 

658 

676 

694 

712 

749 

767 

785 

803 

822 

8 

14.4 

39 

840 

858 

876 

894 

912 

949 

967 

985 

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16.2 

240 

38021 

039 

057 

075 

093 

112 

130 

148 

1  66 

184 

17 

41 

202 

220 

238 

256 

274 

292 

310 

328 

346 

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1.7 

42 

382 

399 

417 

435 

453 

471 

489 

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525 

543 

2 

3.4 

43 

561 

578 

596 

614 

632 

650 

668 

686 

703 

721 

3 

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5.1 

44 

739 

757 

775 

792 

810 

828 

846 

863 

88  1 

899 

4 

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45 

917 

934 

952 

970 

987 

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5 

8-5 

46 

39094 

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129 

146 

164 

182 

199 

217 

235 

252 

6 

IO.2 

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270 

287 

30? 

322 

340 

358 

375 

393 

410 

428 

7 

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48 

445 

463 

480 

498 

515 

533 

55° 

568 

585 

602 

8 

13.6 

49 

620 

637 

655 

672 

690 

707 

724 

742 

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777 

9 

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0 

1 

2 

3 

4 

5 

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7 

8 

9 

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IV 


250-299 


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O 

1 

2 

3   4 

5 

6 

7 

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9 

Pp.  Pts. 

250 

39  794 

811 

829 

846 

863 

88  1 

898 

915 

933 

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18 

51 

967 

985 

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1.8 

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40  140 

157 

175 

192 

209 

226  243 

261 

278 

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2 

3.6 

53 

312 

329 

346 

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398  415 

432 

449 

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c.4 

54 

483 

500 

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535 

552 

569 

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603 

620 

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7-2 

55 

654 

671 

688 

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722 

739  756 

773 

790 

807 

9.0 

56 

824 

841 

858 

875 

892 

909  926 

943 

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976 

6 

10.8 

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7 

12.6 

58 

41  162 

179 

196 

212 

229 

246 

263 

280 

296 

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8 

14.4 

59 

330 

347 

363 

380 

397 

414 

43° 

447 

464 

481 

9 

16.2 

200 

61 

497 
664 

5*4 

68  1 

£ 

547 
714 

564 

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747 

597 
764 

614 
780 

631 

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647 
814 

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1.7 

62 

830 

847 

863 

880 

896 

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929 

946 

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2 

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3  4 

63 

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243 

259  275 

292 

308 

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341 

357 

374 

390 

406 

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8.5 

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488 

5°4 

521 

537 

553 

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586  602 

619 

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6 

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3 

651 

813 

667 
830 

684 
846 

700 
862 

716 

878 

732 
894 

749 
911 

765 
927 

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943 

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13.6 

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9 

15-3 

270 

43  136 

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169 

185 

20  1 

217 

233 

249 

265 

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16 

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297 

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329 

345 

361 

377 

393 

409 

425 

441 

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1.6 

72 

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473 

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505 

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537 

553 

569 

584 

600 

2 

3.2 

73 

616 

632 

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664 

680 

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712 

727 

743 

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3 

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807 

823 

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854 

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933 

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965 

981 

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8.0 

76 

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107 

122 

138 

154 

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217 

232 

6 

9.6 

77 

248 

264 

279 

295 

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326 

342 

358 

373 

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7 

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78 

404 

420 

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45  i 

467 

483 

498 

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529 

545 

8 

12.8 

79 

560 

576 

592 

607 

623 

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654 

669 

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700 

9 

14.4 

280 

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73i 

747 

762 

778 

793 

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824 

840 

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81 

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917 

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82 

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148 

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2 

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179 

194 

209 

225 

240 

255 

271 

286 

301 

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332 

347 

362 

378 

393 

408 

423 

439 

454 

469 

4 

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6.0 

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484 

500 

5*5  530 

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561 

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606 

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7-5 

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637 

652 

667 

682 

697 

712 

728 

743 

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6  !  9.0 

87 

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818 

834 

849 

864  i  879 

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909 

924 

7 

10.5 

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939 

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969 

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015  *030 

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12.0 

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46  090 

105 

120 

135 

150 

165  180 

195 

210 

225 

9 

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290 

240 

255 

270 

285 

300 

3i5 

330 

345 

359 

374 

14 

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389 

404 

419 

434 

449 

464 

479 

494 

5°9 

523 

i 

1.4 

92 

538 

553 

568 

583 

598 

613 

627 

642 

657 

672 

2 

2.8 

93 

687 

702 

7l6 

73i 

746 

761 

776 

790 

805 

820 

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4.2 

94 

835 

850 

864 

879 

894 

909 

923 

938 

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967 

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4 

5.6 

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056 

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144 

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202 

217 

232 

246 

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276 

290 

305 

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349 

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9  12.6 

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1 

2 

3 

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5 

6 

7 

8 

9 

Pp.  Pts. 

300-349 


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0 

1 

2 

3   4 

5 

6 

7 

8   9 

Pp.  Pts. 

300 

47  712 

727 

741 

756  770 

784 

799 

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828 

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857 

871 

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900 

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044 

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101 

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330 

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359 

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544 

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601 

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248 

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276 

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332 

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513 

527 

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568 

582 

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610 

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651 

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721 

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542 

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270 

284 

297 

310 

323 

336 

349 

362 

7 

9.1 

34 

375 

388 

401 

414 

427 

440 

453 

466 

479 

492 

8 

10.4 

35 

5°4 

517 

530 

543 

556 

569 

582 

595 

608 

621 

9 

11.7 

36 

634 

647 

660 

673 

686 

699 

711 

724 

737 

75° 

37 

763 

776 

789 

802 

815 

827 

840 

853 

866 

*879 

38 

892 

901 

917 

930 

943 

956 

969 

982 

994 

12 

39 

53020 

033 

046 

058 

071 

084 

097 

no 

122 

135 

i 

1.2 

340 

148 

161 

173 

1  86 

199 

212 

224 

237 

250 

263 

2   2.4 

41 

275 

288 

301 

314 

326 

339 

352 

364 

377 

390 

3  3-6 

42 
43 

403 

529 

415 
542 

428 
555 

441 
567 

466 
593 

479 
605 

491 
618 

5°4 
631 

III 

4  4.8 
5  6.0 

44 

656 

668 

694 

706 

719 

732 

744 

757 

769 

6  7.2 

45 

782 

794 

807  1  820 

832 

843 

857 

870 

882 

895 

7  8.4 

ft  '   /->  A 

46 

908 

920 

933 

945 

958 

970  983 

995 

*oo8 

*O2O 

o   9.0 

4%  1  »*-v  Q 

47 

54033 

045 

058 

070 

083 

095  108 

1  20 

133 

145 

9  io.o 

48 

158 

170 

183 

195 

208 

220   233 

245 

258 

270 

49 

283 

295 

307 

320 

332 

345 

357 

370 

382 

394 

N. 

0 

1 

2 

3 

4 

5 

6   7 

8 

9 

Pp.  Pts. 

VI 


350-399 


N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Pp.  Pts. 

350 

54407 

419 

432 

444 

456 

469 

481 

494 

506 

518 

51 

53i 

543 

555 

568 

580 

593 

605 

617 

630 

642 

52 

654 

667 

679 

691 

704 

716 

728 

741 

753 

765 

53 

777 

790 

802 

814 

827 

839 

851 

864 

876 

888 

13 

54 

900 

913 

92? 

937 

949 

962 

974 

986 

998 

*on 

i 

2 

*-3 

2.6 

55 

55°23 

035 

047 

060 

072 

084 

096 

1  08 

121 

133 

7 

•2.Q 

56 

143 

157 

169 

182 

194 

206 

218 

230 

242 

255 

O 

4 

O  s 
C.2 

57 

267 

279 

291 

303 

315 

328 

340 

352 

364 

376 

6.5 

58 

388 

400 

4i3 

425 

437 

449 

461 

473 

485 

497 

^ 

59 

509 

522 

534 

546 

558 

570 

582 

594 

606 

618 

7 

9.1 

360 

630 

642 

654 

666 

678 

691 

7°3 

7i5 

727 

739 

8 

IO.4 

61 

75  i 

763 

775 

787 

799 

811 

823 

831 

847 

859 

9 

II.7 

62 

871 

883 

895 

907 

919 

93  i 

943 

955 

967 

979 

63 

991 

*oo3 

*oi5 

*O27 

*o38 

*O5O 

*062 

*074 

*o86 

*098 

64 

56  no 

122 

134 

146 

158 

170 

182 

194 

205 

217 

12 

65 

229 

241 

253 

265 

277 

289 

301 

312 

324 

336 

i 

1.2 

66 

348 

360 

372 

384 

396 

407 

419 

43i 

443 

451 

2 

2.4 

67 

467 

478 

490 

502 

5'4 

526 

538 

549 

561 

573 

1 

T^ 

3.6 

68 

585 

597 

608 

620 

632 

644 

656 

667 

679 

691 

O 

4 

i8 

69 

7°3 

714 

726 

738 

750 

761 

773 

785 

797 

808 

5 

6.0 

370 

820 

832 

844 

855 

867 

879 

891 

902 

914 

926 

6 

7-2 

7i 

937 

949 

961 

972 

984 

996 

*oo8 

*oi9 

*03i 

*043 

7 

8.4 

72 

57054 

066 

078 

089 

101 

"3 

124 

136 

148 

I59 

8 

9'« 

73 

171 

183 

194 

206 

217 

229 

241 

252 

.264 

276 

9 

10.8 

74 

287 

299 

310 

322 

334 

345 

357 

368 

380 

392 

75 

403 

415 

426 

438 

449 

461 

473 

484 

496 

5°7 

76 

519 

530 

542 

553 

565 

576 

588 

600 

611 

623 

ii 

77 

634 

646 

657 

669 

680 

692 

7°3 

7J5 

726 

738 

i 

I.I 

78 

749 

761 

772 

784 

795 

807 

818 

830 

841 

852 

2 

2.2 

79 

864 

875 

887 

898 

910 

921 

933 

944 

955 

967 

3 

3-3 

380 

978 

990 

*OOI 

*oi3 

*024 

*035 

*047 

"=058 

"•070 

*o8i 

4 

4-4 

81 

58092 

104 

"5 

127 

138 

149 

161 

172 

184 

195 

5 

5-5 

82 

206 

218 

229 

240 

252 

263 

274 

286 

297 

309 

6 

6.6 

83 

320 

331 

343 

354 

365 

377 

388 

399 

410 

422 

7 

7'7 

84 

433 

444 

456 

467 

478 

490 

5oi 

512 

524 

535 

8 

8.8 

85 

546 

557 

569 

580 

591 

602 

614 

625 

636 

647 

9 

9.9 

86 

659 

670 

68  1 

692 

704 

7i5 

726 

737 

749 

760 

87 

782 

794 

805 

816 

827 

838 

850 

86  1 

872 

88 

883 

894 

906 

917 

928 

939 

95° 

961 

973 

984 

10 

89 

991 

*oo6 

*oi7 

*028 

*040 

*o5i 

*062 

*073 

*o84 

*095 

i 

I.O 

390 

59  106 

118 

129 

140 

J51 

162 

J73 

184 

J95 

207 

2 

2.0 

91 

218 

229 

240 

25  1 

262 

273 

284 

295 

306 

3i8 

3 

3-o 

92 

329 

340 

351 

362 

373 

384 

395 

406 

417 

428 

4 

4.0 

93 

439 

45° 

461 

472 

483 

494 

506 

5J7 

528 

539 

5 

5-o 

94 

550 

561 

572 

583 

594 

605 

616 

627 

638 

649 

6 

6.0 

95 

660 

671 

682 

693 

704 

7*5 

726 

737 

748 

759 

g 

7.0 
8.0 

96 

770 

780 

791 

802 

813 

824 

83i 

846 

857 

868 

o.o 

97 

879 

890 

901 

912 

923 

934 

945 

956 

966 

977 

^•w 

98 

988 

999 

*OIO 

*O2I 

*O32 

*Q43 

*Q54 

*o65 

*076 

*o86 

99 

60  097 

108 

119 

I30 

141 

152 

163 

173 

184 

195 

N. 

O 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Pp.  Pts. 

400-449 


VII 


N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Pp.  Pts. 

400 

60  206 

217 

228 

239 

249 

260 

271 

282 

293 

304 

01 

3H 

325 

336 

347 

358 

369 

379 

390 

401 

412 

02 

423 

433 

444 

455 

466 

477 

487 

498 

5°9 

520 

03 
04 

53i 
638 

649 

IS 

563 
670 

574 
681 

£ 

595 
7°3 

606 
713 

617 
724 

627 
735 

05 

746 

756 

767 

778 

788 

799 

810 

821 

831 

842 

06 

853 

863 

874 

885 

895 

906 

917 

927 

938 

949 

ii 

07 

959 

97° 

981 

991 

*OO2 

013 

*023 

*034 

*o45 

*055 

i 

i.i 

08 

6  1  066 

077 

087 

098 

109 

119 

130 

140 

1S1 

162 

2 

2.2 

09 

172 

183 

194 

204 

215 

225 

236 

247 

257 

268 

3 

3-3 

410 

278 

289 

300 

310 

32I 

331 

342 

352 

363 

374 

4 

4.4 

ii 

384 

395 

405 

416 

426 

437 

448 

458 

469 

479 

I 

66 

12 

490 

500 

511 

521 

S32 

542 

553 

563 

574 

584 

o 

o.o 

13 

595 

606 

616 

627 

637 

648 

658 

669 

679 

690 

I 

7-7 
8.8 

14 

700 

711 

721 

73i 

742 

752 

763 

773 

784 

794 

15 

805 

815 

826 

836 

847 

857 

868 

878 

888 

899 

9 

9.9 

16 

909 

920 

930 

941 

95  * 

962 

972 

982 

993 

*cx>3 

17 

62  014 

024 

034 

045 

°55 

066 

076 

086 

097 

107 

18 

118 

128 

138 

149 

1S9 

170 

1  80 

190 

20  1 

211 

19 

221 

232 

242 

252 

263 

273 

284 

294 

3°4 

315 

420 

321 

335 

346 

356 

366 

377 

387 

397 

408 

418 

10 

21 

428 

439 

449 

459 

469 

480 

490 

500 

5" 

521 

I(-V 

22 

531 

542 

552 

562 

572 

583 

593 

603 

613 

624 

2 

,\J 

2  O 

23 

634 

644 

653 

665 

675 

685 

696 

706 

716 

726 

•7  /-k 

24 

737 

747 

757 

767 

778 

788 

798 

808 

818 

829 

4 

3-° 
4.0 

25 

839 

849 

859 

870 

880 

890 

900 

910 

921 

931 

5-° 

26 

941 

95  * 

961 

972 

982 

992 

*002 

*OI2 

*O22 

*033 

6 

6.0 

27 

63043 

053 

063 

073 

083 

094 

104 

114 

124 

134 

7 

7.0 

28 

144 

155 

165 

175 

185 

195 

205 

215 

225 

*236 

8 

8.0 

29 

246 

256 

266 

276 

286 

296 

306 

317 

327 

337 

9 

9.0 

430 

347 

357 

367 

377 

387 

397 

407 

417 

428 

438 

3i 

448 

458 

468 

478 

488 

498 

508 

5*8 

528 

538 

32 

548 

558 

568 

579 

589 

599 

609 

619 

629 

639 

33 

649 

659 

669 

679 

689 

699 

709 

719 

729 

739 

34 

749 

759 

769 

779 

789 

799 

809 

819 

829 

839 

35 

849 

859 

869 

879 

889 

899 

909 

919 

929 

939 

9 

36 

949 

959 

969 

979 

988 

998 

*oo8 

*oi8 

*028 

*038 

i 

0.9 

37 

64048 

058 

068 

078 

088 

098 

108 

118 

128 

137 

2 

1.8 

3^ 

147 

:57 

167 

177 

187 

197 

207 

217 

227 

237 

3 

2.7 

39 

246 

256 

266 

276 

286 

296 

306 

3i6 

326 

335 

4 

3-6 

440 

345 

355 

365 

375 

385 

395 

404 

414 

424 

434 

5 

4-5 

r  A 

4i 

444 

454 

464 

473 

483 

493 

503 

5i3 

523 

S32 

iv4 

42 

542 

552 

562 

572 

582 

59i 

60  1 

611 

621 

631 

7 

6'3 

hy  /> 

43 

640 

650 

660 

670 

680 

689 

699 

709 

719 

729 

81 

44 

738 

748 

758 

768 

777 

787 

797 

807 

816 

826 

9 

0.  1 

45 

836 

846 

856 

865 

875 

885 

895 

904 

914 

924 

46 

.933 

943 

953 

963 

972 

982 

992 

*OO2 

*OII 

*O2I 

47 

65031 

040 

050 

060 

070 

079 

089 

099 

108 

118 

48 

128 

137 

147 

157 

167 

176 

1  86 

196 

205 

215 

49 

225 

234 

244 

254 

263 

273 

283 

292 

302 

312 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Pp.  Pts. 

450-499 


N. 

0 

1 

2 

3 

4 

5 

<5 

7 

8 

9 

Pp.  Pts. 

450 

65321 

331 

341 

350 

360 

369 

379 

389 

398 

408 

51 

418 

427 

437 

447 

456 

466 

475 

485 

495 

5°4 

52 

5H 

523 

533 

543 

552 

562 

571 

58i 

59i 

600 

53 

610 

619 

629 

639 

648 

658 

667 

677 

686 

696 

54 

706 

7*5 

725 

734 

744 

753 

763 

772 

782 

792 

55 

801 

811 

820 

830 

839 

849 

858 

868 

877 

887 

10 

56 

896 

906 

916 

925 

935 

944 

954 

963 

973 

982 

i 

I.O 

57 

992 

*OOI 

*on 

*020 

*O3O 

*039 

*o49 

*o58 

*o68 

*°77 

2 

2.O 

58 

66087 

096 

106 

»5 

124 

134 

H3 

J53 

162 

172 

3 

3-0 

59 

181 

191 

200 

2IO 

219 

229 

238 

247 

257 

266 

4 

4.0 

460 

276 

285 

295 

3°4 

3H 

323 

332 

342 

35i 

36i 

| 

5-° 
6.0 

61 

370 

380 

389 

398 

408 

4i7 

427 

436 

445 

455 

7 

7.0 

62 

464 

474 

483 

492 

502 

5" 

521 

530 

539 

549 

8 

8.0 

63 
64 

558 
652 

567 
661 

577 
671 

586 
680 

596 
689 

605 
699 

614 

708 

624 
717 

633 
727 

642 

736 

9 

9.0 

65 

745 

755 

764 

773 

783 

792 

801 

811 

820 

829 

66 

839 

848 

857 

867 

876 

885 

894 

904  913 

922 

67 

932 

941 

95° 

960 

969 

978 

987 

997 

*oo6 

*oi5 

68 

67025 

034 

043 

052 

062 

071 

080 

089 

099 

108 

69 

117 

127 

136 

H5 

'54 

164 

173 

182  j  191 

20  1 

470 

210 

219 

228 

237 

247 

256 

265 

274  284!  293 

7i 

302 

3ii 

321 

33° 

339 

348 

357 

367  376 

3^5 

i 

0.9 

72 

394 

403 

4i3 

422 

43i 

440 

449 

459  468 

477 

2 

1.8 

73 

486 

495 

504 

5*4 

523 

532 

54i 

55°:  56o 

569 

7 

2.7 

74 

578 

587 

596 

605 

614 

624 

633 

642  i  651 

660 

O 

4 

/ 

3-6 

75 

669 

679 

688 

697 

706 

7i5 

724 

733  742 

752 

5 

4-5 

76 

761 

770 

779 

788  797 

806 

815 

825  834 

843 

6 

5-4 

77 

852 

861 

870 

879 

888 

897 

906 

916  925 

934 

7 

6-3 

78 

943 

952 

961 

97° 

979 

988 

997 

*oo6 

*oi5  *O24 

8 

7-2 

79 

68034 

043 

052 

06  1 

070 

079 

088 

097 

1  06 

"5 

9 

8.1 

480 

124 

133 

142 

151 

1  60 

169 

178 

187 

196 

205 

81 

215 

224 

233 

242 

251 

260 

269 

278  :  287   296 

82 

305 

3H 

323 

332 

34i 

350 

359 

368  I  377  386 

83 

395 

404 

4i3 

422 

43  1 

440 

449 

458  i  467  476 

84 

485 

494 

502 

5" 

520 

529 

538 

547 

556:  565 

85 

574 

583 

592 

60  1 

610 

619 

628 

637 

646  655 

86 

664 

673 

681 

690 

699 

708 

717 

726 

735  i  744 

8 

87 

753 

762 

771 

780 

789 

797 

806 

815 

824 

833 

i 

0.8 

88 

842 

851 

860 

869 

878 

886 

895 

904 

9i3 

922 

2 

1.6 

89 

931 

940 

949 

958 

966 

975 

984 

993 

*OO2 

*OII 

3 

2.4 

490 

69  020 

028 

037 

046 

°55 

064 

°73 

082 

090 

099 

4 

3-2 

9i 

108 

117 

126 

i35 

144 

1S2 

161 

170 

I79 

1  88 

5 

4.0 

.  o 

92 

197 

205 

214 

223 

232 

241 

249 

258 

267 

276 

° 

4.0 

e  f\ 

93 

285 

294 

302 

3" 

320 

329 

338 

346 

355 

364 

7 

I 

6  A 

94 

373 

38i 

390 

399 

408 

4^7 

425 

434 

443 

452 

9 

0.4 

7.2 

95 

461 

469 

478 

487 

496 

5°4 

5'3 

522 

53i 

539 

96 

548 

557 

566 

574 

583 

592 

601 

609 

618 

627 

97 

636 

644 

653 

662 

671 

679 

688 

697 

7°5 

714 

98 

723 

732 

740 

749 

758 

767 

775 

784 

793 

801 

99 

810 

819 

827 

836 

84! 

854 

862 

871  880 

888 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Pp.  Pts. 

500-549 


IX 


N. 

O 

1 

2 

3 

4 

5 

6 

7  |  8 

9 

Pp.  Pts. 

500 

69897 

906 

914 

923 

932 

940 

949 

958 

966 

975 

01 

984 

992 

*OOI 

*OIO 

*oi8 

*027 

"036 

*044 

*°53 

*062 

02 

70  070 

079 

088 

096 

105 

114 

122 

131 

140 

148 

°3 

J57 

165 

*74 

183 

191 

2OO 

209 

217 

226 

234 

04 

243 

252 

260 

269 

278 

286 

293 

3°3 

312 

321 

05 

329 

338 

346 

353 

364 

372 

38i 

389 

398 

406 

9 

06 

4i5 

424 

432 

441 

449 

458 

467 

475 

484 

492 

i 

?8 

07 

08 

501 
586 

5°9 
593 

518 
603 

526 
612 

535 
621 

544 
629. 

1% 

561 
646 

% 

578 
663 

2 

3 

I.o 

2'7 

09 

672 

680 

689 

697 

706 

714 

723 

73i 

740 

749 

4 

3-6 

510 

757 

766 

774 

783 

791 

800 

808 

817 

825 

834 

i 

4-5 

c.4 

ii 

842 

851 

859 

868 

876 

885 

893 

902 

910 

919 

J   i 
6.3 

12 

927 

935 

944 

952 

961 

969 

978 

986 

993 

*003 

g 

O 

7.2 

13 

71  OI2 

020 

029 

037 

046 

054 

063 

071 

079 

088 

9 

8.1 

14 

096 

103 

H3 

122 

130 

•39 

147 

'55 

164 

172 

15 

181 

189 

198 

206 

214 

223 

231 

240 

248 

257 

16 

265 

273 

282 

290 

299 

307 

315 

324 

332 

34i 

J7 

349 

357 

366 

374 

383 

39i 

399 

408 

416 

423 

18 

433 

441 

4jo 

458 

466 

475 

483 

492 

500 

508 

19 

5J7 

525 

533 

542 

550 

559 

567 

575 

584 

592 

520 

600 

609 

617 

625 

634 

642 

650 

659 

667 

•675 

8 

21 

684 

692 

700 

709 

717 

725 

734 

742 

75° 

759 

j 

0.8 

22 

767 

784 

792 

800 

809 

817 

825 

834 

842 

2 

1.6 

23 

850 

858 

867 

875 

883 

892 

900 

908 

917 

923 

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2.4 

24 

933 

941 

95° 

958 

966 

975 

983 

991 

999 

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4 

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3-2 

25 

72  016 

024 

032 

041 

049 

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066 

074 

082 

090 

5 

4.0 

26 

099 

107 

115 

123 

132 

140 

148 

156 

165 

173 

6 

4-8 

27 

181 

189 

198 

206 

214 

222 

230 

^39 

247 

255 

7 

5.6 

28 

263 

272 

280 

288 

296 

3°4 

313 

321 

329 

337 

8 

6-4 

29 

346 

354 

362 

370 

378 

387 

395 

403 

411 

419 

9 

7.2 

530 

428 

436 

444 

452 

460 

469 

477 

485 

493 

5oi 

3i 

5°9 

518 

526 

534 

542 

550 

558 

567 

575 

583 

32 

59i 

599 

607 

616 

624 

632 

640 

648 

656 

665 

33 

673 

681 

689 

697 

705 

713 

722 

730 

738 

746 

34. 

754 

762 

770 

779 

787 

795 

803 

811 

819 

827 

35 

835 

843 

852 

860 

868 

876 

884 

892 

900 

908 

36 

916 

925 

933 

941 

949 

957 

965 

973 

981 

989 

7 

37 

997 

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i 

0.7 

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73078 

086 

094 

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in 

119 

127 

133 

143 

151 

2 

1.4 

39 

159 

167 

175 

183 

191 

199 

207 

215 

223 

231 

3 

2.1 

540 

239 

247 

255 

263 

272 

280 

288 

296 

3°4 

312 

4 

2.8 

41 

320 

328 

336 

344 

352 

360 

368 

376 

384 

392 

£ 

3-5 

42 

400 

408 

416 

424 

432 

440 

448 

456 

464 

472 

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4.2 

43 

480 

488 

496 

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512 

520 

528 

536 

544 

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7 

4-9 

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44 

560 

568 

576 

584 

592 

600 

608 

616 

624 

632 

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6  3 

45 

640 

648 

656 

664 

672 

679 

687 

695 

703 

711 

w»J 

46 

719 

727 

735 

743 

751 

759 

767 

773 

783 

791 

47 

799 

807 

815 

823 

830 

838 

846 

854 

862 

870 

48 

878 

886 

894 

902 

910 

918 

926 

933 

941 

949 

49 

957 

965 

973 

981 

989 

997  "005 

*oi3 

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N. 

O    1 

2   3 

4 

5 

6   7 

8 

9 

Pp.  Pts. 

550-599 


N. 

0 

1  |  2 

3 

4 

5 

6   7 

8 

9 

Pp.  Pts. 

550 

74036 

044 

052 

060 

068 

076 

084 

092 

099 

107 

51 

"5 

123 

»3i 

139 

147 

155 

162 

170 

178 

186 

52 

194 

202 

210 

218 

225 

233 

241 

249 

257 

265 

53 

273 

280 

288 

296 

304 

312 

320 

327 

335 

343 

54 

35i 

359 

367 

374 

382 

390 

398 

406 

414 

421 

55 

429 

437 

445 

453 

461 

468 

476 

484 

492 

500 

56 

5°7 

5'5 

523 

53i 

539 

547 

554 

562 

57° 

578 

57 

586 

593 

60  1 

609 

617 

624 

632 

640 

648 

656 

58 

663 

671 

679 

687 

695 

702 

710 

718 

726 

733 

59 

741 

749 

757 

764 

772 

780 

788 

796 

803 

811 

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819 

827 

834 

842 

850 

858 

865 

873 

88  1 

889 

8 

T   08 

61 

896 

904 

912 

920 

927 

935 

943 

95° 

958 

966 

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62 

974 

981 

989 

997 

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63 

75051 

059 

066 

074 

082 

089 

097 

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1  20 

3  2.4 

64 

128 

136 

H3 

151 

'59 

1  66 

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182 

189 

197 

4  3-2 

C   A  O 

65 

205 

213 

220 

228 

236 

243 

25  * 

259 

266 

274 

j  '!••*•' 

6  4.8 

66 

282 

289 

297 

305 

312 

320 

328 

335 

343 

35  i 

7  5-6 

67 

358 

366 

374 

38i 

389 

397 

404 

412 

420 

427 

68 

431 

442 

45° 

458 

465 

473 

481 

488 

406 

5°4 

91  7-2 

69 

511 

5'9 

526 

534 

542 

549 

557 

565 

572 

580 

570 

587 

595 

603 

610 

618 

626 

633 

641 

648 

656 

71 

664 

671 

679 

686 

694 

702 

709 

717 

724 

732 

72 

740 

747 

75? 

762 

770 

778 

785 

800 

808 

73 

815 

823 

831 

838 

846 

853 

861 

868 

876 

884 

74 

891 

899 

906 

914 

921 

929 

937 

944 

952 

959 

75 

967 

974 

982 

989 

997 

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76 

76  042 

050 

057 

065 

072 

080 

087 

095 

103 

no 

77 

118 

125 

133 

140 

148 

155 

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170 

178 

185 

78 

193 

200 

208 

215 

223 

230 

238 

245 

253 

260 

79 

268 

275 

283 

290 

298 

305 

313 

320 

328 

335 

580 

343 

350 

358 

365 

373 

380 

388 

395 

403 

410 

81 

418 

425 

433 

440 

448 

455 

462 

47° 

477 

485 

82 

492 

500 

507 

515 

522 

530 

537 

545 

552 

559 

I  O  7 

83 

567 

574 

582 

589 

597 

604 

612 

619 

626 

634 

i  «-»./ 

O    1  A 

84 

641 

649 

656 

664 

671 

678 

686 

693 

701 

708 

*   1-4 

3  2.1 

85 

716 

723 

730 

738 

745 

753 

760 

768 

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782 

4  2.8 

86 

790 

797 

805 

812 

819 

827 

834 

842 

849 

856 

5  3-5 

87 

864 

871 

879 

886 

893 

901 

908 

916 

923 

930 

6  4.2 

88 

938 

945 

953 

960 

967 

975 

982 

989 

997 

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7  4-9 

89 

77012 

019 

026 

034 

041 

048 

056 

063 

070 

078 

8  5.6 

590 

085 

093 

IOO 

107 

H5 

122 

129 

137 

144 

151 

9  6.3 

9i 

159 

1  66 

173 

181 

1  88 

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203 

2IO 

217 

225 

92 

232 

240 

247 

254 

262 

269 

276 

283 

291 

298 

93 

305 

313 

320 

327 

335 

342 

349 

357 

364 

37i 

94 

379 

386 

393 

401 

408 

415 

422 

43° 

437 

444 

95 

452 

459 

466 

474 

481 

488 

495 

5°3 

510 

5'7 

96 

525 

532 

539 

546 

554 

561 

568 

576 

583 

590 

97 

597 

605 

612 

619 

627 

634 

641 

648 

656 

663 

98 

670 

677 

685 

692 

699 

706 

7H 

721 

728 

735 

99 

743 

750 

757 

764 

772 

779 

786 

793 

80  1 

808 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Pp.  Pts. 

600-649 


N. 

0 

1 

2 

3 

4 

5 

6   7 

8 

9 

Pp.  Pts. 

000 

77815 

822 

830 

837 

844 

851 

859 

866 

873 

880 

01 

887 

895 

902 

909 

916 

924 

931 

938 

945 

952 

02 

960 

967 

974 

981 

988 

996 

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03 

78032 

039 

046 

053 

061 

068 

075 

082 

089 

097 

04 

104 

in 

118 

125 

132 

140 

147 

154 

161 

1  68 

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176 

183 

190 

197 

204 

211 

219 

226 

233 

240 

8 

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06 

247 

254 

262 

269 

276 

283 

290 

297 

305 

312 

i 

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07 

319 

326 

333 

340 

347 

355 

362 

369 

376 

383 

2 

I.O 

08 

390 

398 

405 

412 

419 

426 

433 

440 

447 

455 

3 

2-4 

09 

462 

469 

476 

483 

490 

497 

5°4 

512 

519 

526 

4 

3-2 

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610 

533 

540 

547 

554 

561 

569 

576 

583 

590 

597 

5 

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4.8 

ii 

604 

611 

618 

625 

633 

640 

647 

654 

661 

668 

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5.6 

12 

675 

682 

689 

696 

704 

711 

718 

725 

732 

739 

g 

6.4 

13 

746 

753 

760 

767 

774 

781 

789 

796 

803 

810 

9 

7.2 

14 

817 

824 

831 

838 

845 

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859 

866 

873 

880 

15 

888 

895 

902 

909 

916 

923 

930 

937 

944 

95  1 

16 

958 

965 

972 

979 

986 

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043 

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057 

064 

071 

078 

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092 

18 

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106 

1^3 

120 

127 

134 

141 

148 

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162 

19 

169 

176 

183 

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197 

204 

211 

218 

225 

232 

620 

239 

246 

253 

260 

267 

274 

28l 

288 

295 

302 

7 

21 

309 

3i6 

323 

330 

337 

344 

351 

358 

365 

372 

i 

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0.7 

22 

379 

386 

393 

4OO 

407 

414 

42I 

428 

435 

442 

2 

1.4 

23 

449 

456 

463 

47° 

477 

484 

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498 

505 

511 

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2.1 

24 

518 

525 

532 

539 

546 

553 

560 

567 

574 

581 

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4 

2.8 

25 

588 

591 

602 

609 

616 

623 

630 

637 

644 

650 

5 

3-5 

26 

657 

664 

671 

678 

685 

692 

699 

706 

7*3 

720 

6 

4.2 

27 

727 

734 

741 

748 

754 

761 

768 

775 

782 

789 

7 

4.9 

28 

796 

803 

810 

817 

824 

831 

837 

844 

851 

858 

8 

5.6 

29 

865 

872 

879 

886 

893 

900 

906 

9U 

920 

927 

9 

6-3 

630 

934 

941 

948 

955 

962 

969 

975 

982 

989 

996 

31 

80  003 

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017 

024 

030 

037 

044 

05  * 

058 

065 

32 

072 

079 

085 

092 

099 

106 

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1  20 

127 

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33 

140 

147 

154 

161 

1  68 

175 

182 

1  88 

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202 

34 

209 

216 

223 

229 

236 

243 

250 

257 

264 

271 

35 

277 

284 

291 

298 

305 

312 

3i8 

325 

332 

339 

36 

346 

353 

359 

366 

373 

380 

387 

393 

400 

407 

6 

37 

414 

421 

428 

434 

441 

448 

455 

462 

468 

475 

i 

0.6 

38 

482 

489 

496 

502 

5°9 

516 

523 

530 

536 

543 

2 

1.2 

39 

55° 

557 

564 

570 

577 

584 

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598 

604 

611 

3 

1.8 

640 

618 

625 

632 

638 

645 

652 

659 

665 

672 

679 

4 

2.4 

41 

686 

693 

699 

706 

7J3 

720 

726 

733 

740 

747 

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3-° 

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42 

754 

760 

767 

774 

781 

787 

794 

80  1 

808 

814 

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43 

821 

828 

835 

841 

848 

855 

862 

868 

875 

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4-2 

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44 

889 

895 

902 

909 

916 

922 

929 

936 

943 

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9 

4.5 
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45 

956 

963 

969 

976 

983 

990 

996 

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81  023 

030 

037 

043 

050 

057 

064 

070 

077 

084 

47 

090 

097 

104 

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117 

124 

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137 

144 

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48 

158 

164 

171 

178 

184 

191 

198 

204 

211 

218 

49 

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231 

238 

245 

251 

258 

265 

271 

278 

285 

N. 

0 

1 

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3   4 

5   6 

7 

8   9 

Pp.  Pts. 

Xll 


650-699 


N. 

0 

1  |  2 

3 

4 

5 

6 

7 

8 

9 

Pp.  Pts. 

650 

81  291 

298 

30? 

311 

3i8 

32? 

33i 

338 

34? 

351 

51 

358 

36? 

378 

38? 

398 

405 

411 

418 

52 

425 

431 

438 

44? 

458 

471 

478 

48? 

53 

491 

498 

505 

5" 

518 

52? 

53' 

538 

544 

551 

54 

558 

564 

578 

584 

598 

604 

611 

617 

55 

624 

631 

637 

644 

651 

657 

664 

671 

677  684 

56 

690 

697 

704 

710 

717 

723 

730 

737 

743  75° 

57 

757 

763 

770 

776 

783 

790 

796 

803 

809  816 

58 

823 

829 

836 

842 

849 

856 

862 

869 

875 

882 

59 

889 

895 

902 

908 

9i? 

921 

928 

93? 

941 

948 

660 

954 

961 

968 

974 

981 

987 

994 

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7 

r\  *r 

61 

82  020 

027 

033 

040 

046 

053 

060 

066 

073 

079 

0.7 

62 

086 

092 

099 

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112 

119 

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132 

138 

H5 

• 

1.4 

63 

151 

158 

164 

171 

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184 

191 

197 

204 

2IO 

3 

2.  1 

64 

217 

223 

230 

236 

243 

249 

256 

263 

269 

276 

5 

3-5 

65 

282 

289 

295 

302 

308 

315 

321 

328 

334 

341 

6 

4.2 

66 

347 

354 

360 

373 

380 

387 

393 

400 

406 

7 

4.9 

67 

413 

419 

426 

432 

439 

445 

452 

458 

46? 

471 

8 

5-6 

68 

478 

484 

491 

497 

5°4 

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5X7 

523 

53° 

536 

9 

6.3 

69 

543 

549 

556 

562 

569 

575 

582 

588 

59? 

60  1 

670 

607 

614 

620 

627 

633 

640 

646 

653 

659 

666 

71 

672 

679 

685 

692 

698 

70? 

711 

718 

724 

73° 

72 

737 

743 

75° 

756 

763 

769 

776 

782 

789 

795 

73 

802 

808 

814 

821 

827 

834 

840 

847 

853 

860 

74 

866 

872 

879 

885 

892 

898 

905 

911 

918 

924 

75 

93° 

937 

943 

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963 

969 

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76 

995 

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77 

83059 

065 

072 

078 

08? 

091 

097 

104 

I  10 

117 

78 

123 

129 

136 

142 

149 

155 

161 

1  68 

174 

181 

79 

187 

193 

200 

206 

2I3 

219 

225 

232 

238 

24? 

680 

25  1 

257 

264 

270 

276 

283 

289 

296 

302 

308 

81 

3I5 

321 

327 

334 

340 

347 

353 

359 

366 

372 

6 

82 

378 

385 

398 

404 

410 

423 

429 

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0.6 

83 

442 

448 

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461 

467 

474 

480 

487 

493 

499 

2 

1.2 

84 

506 

512 

518 

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544 

550 

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3 

1.8 

85 

569 

575 

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601 

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613 

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4 

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632 

639 

651 

658 

664 

670 

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5 

3.0 

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702 

708 

721 

727 

734 

740 

746 

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6 

3-6 

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55 

759 

765 

771 

778 

784 

790 

797 

803 

809 

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7 

4.2 

89 

822 

828 

835 

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847 

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866 

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8 

4.8 

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897 

904 

910 

916 

923 

929 

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54 

91 

948 

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960 

967 

973 

979 

985 

992 

998 

92 

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017 

023 

029 

036 

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061 

067 

93 

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080 

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092 

098 

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117 

123 

130 

94 

136 

142 

148 

155 

161 

167 

173 

180 

1  86 

192 

95 

198 

205 

211 

217 

223 

230 

236 

242 

248 

25? 

96 

261 

267 

273 

280 

286 

292 

298 

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317 

97 

323 

330 

336 

342 

348 

354 

361 

367 

373 

379 

98 

386 

392 

398 

404 

410 

417 

423 

429 

435 

442 

99 

448 

454 

460 

466 

473 

479 

48? 

491 

497 

504 

N. 

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1 

2 

3 

4 

5  |  6 

7   8 

9 

Pp.  Pts. 

700-749 


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N. 

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1 

2 

3 

4 

5 

6 

7 

8 

9 

Pp.  Pts. 

700 

84  510 

516 

522 

528 

533 

541 

547 

553 

559 

566 

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572 

578 

584 

590 

597 

603 

609 

615 

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652 

658 

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671 

677 

683 

689 

03 

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702 

708 

714 

720 

726 

733 

739 

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04 

757 

763 

770 

776 

782 

788 

794 

800 

807 

05 

819 

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831 

837 

844 

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856 

862 

868 

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7 

06 

880 

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899 

905 

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917 

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0.7 

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973 

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2 

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016 

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040 

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3 

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09 

065 

071 

077 

083 

089 

095 

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107 

114 

1  20 

4 

2.8 

710 

126 

132 

138 

144 

150 

156 

163 

169 

173 

181 

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3-5 
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187 

193 

199 

205 

211 

217 

224 

230 

236 

242 

7 

4.9 

12 

248 

254 

260 

266 

272 

278 

283 

291 

297 

303 

8 

5.6 

14 

309 
370 

376 

321 

382 

388 

333 
394 

339 
400 

406 

352 
412 

358 
418 

364 

423 

9 

6-3 

15 

43  1 

437 

443 

449 

455 

461 

467 

473 

479 

485 

16 

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497 

5°3 

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522 

528 

534 

540 

546 

17 

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558 

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570 

576 

582 

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594 

600 

606 

18 

612 

618 

625 

631 

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643 

649 

653 

66  1 

667 

19 

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691 

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7°3 

709 

721 

727 

720 

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112 

118 

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130 

136 

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147 

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165 

171 

177 

183 

189 

195 

201 

207 

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28 

213 

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225 

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29 

273 

279 

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308 

320 

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54 

730 

332 

338 

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350 

356 

362 

368 

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380 

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31 

392 

398 

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433 

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33 

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528 

473 
534 

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546 

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570 

576 

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599 

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617 

623 

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629 

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641 

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652 

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36 

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700 

705 

711 

717 

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729 

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11 

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806 

753 
812 

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835 

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788 
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794 
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800 
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864 

870 

876 

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900 

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911 

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740 

923 

929 

933 

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221 

227 

233 

239 

243 

251 

256 

262 

268 

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274 

280 

286 

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297 

303 

309 

313 

320 

326 

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332 

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349 

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361 

367 

373 

379 

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48 

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396 

402 

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419 

425 

437 

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49 

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489 

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500 

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750-799 


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6 

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523 

529 

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547 

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564 

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576 

581 

587 

593 

599 

604 

610 

616 

52 

622 

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633 

639 

645 

651 

656 

662 

668 

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53 

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685 

691 

697 

703 

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714 

720 

726 

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754 

760 

766 

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777 

783 

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812 

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846 

56 

852 

858 

864 

869 

875 

88  1 

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892 

898 

904 

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915 

921 

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938 

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955 

961 

58 

967 

973 

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156 

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167 

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213 

218 

224 

230 

235 

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247 

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275 

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610 

615 

621 

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632 

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643 

770 

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655 

660 

666 

672 

677 

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689 

694 

700 

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705 

711 

717 

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750 

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773 

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812 

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75 

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131 

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165 

170 

176 

182 

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193 

198 

204 

780 

209 

215 

221 

226 

232 

237 

243 

248 

254 

260 

81 

265 

271 

276 

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287 

293 

298 

304 

310 

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326 

332 

337 

343 

348 

354 

360 

365 

371 

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83 

376 

382 

387 

393 

398 

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409 

415 

421 

426 

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432 

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625 

631 

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763 

768 

774 

779 

785 

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796 

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856 

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900 

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108 

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119 

124 

129 

135 

140 

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146 

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162 

1  68 

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179 

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98 

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211 

217 

222 

227 

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244 

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99 

255 

260 

266 

271 

276 

282 

287 

293 

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0 

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7 

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9 

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800-849 


XV 


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0    1 

2 

3 

4 

5 

6 

7 

8 

9 

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800 

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320 

325 

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336 

342 

347 

352 

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01 

363 

369 

374 

380 

385 

390 

396 

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07 

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693 

698 

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714 

720 

725 

730 

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752 

757 

763 

768 

773 

779 

784 

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800 

806 

811 

816 

822 

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849 

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859 

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902 

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913 

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100 

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116 

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126 

132 

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3-6 

16 

169 

174 

1  80 

190 

196 

201 

206 

212 

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228 

233 

238 

243 

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254 

259 

263 

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8 

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18 

275 

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286 

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297 

302 

307 

312 

318 

323 

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6.4 

19 

328 

334 

339 

344 

350 

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360 

365 

371 

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381 

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392 

397 

403 

408 

413 

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429 

21 

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22 

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23 

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545 

551 

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566 

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577 

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24 

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598 

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630 

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25 

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66  1 

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693 

26 

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709 

714 

719 

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735 

740 

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27 

751 

756 

761 

766 

772 

777 

782 

787 

793 

798 

28 

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808 

814 

819 

824 

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834 

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845 

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29 

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929 

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169 

174 

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247 

252 

257 

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273 

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314 

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324 

330 

335 

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392 

397 

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511 

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547 

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763 
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202 

207 

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217 

222 

227 

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258 

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187 

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274 

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328 

332 

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357 

361 

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450 

454 

459 

464 

468 

22 

473 

478 

483 

487 

492 

497 

501 

506 

5" 

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23 

520 

525 

530 

534 

539 

544 

548 

553 

558 

562 

24 

567 

572 

577 

58i 

586 

59i 

595 

600 

605 

609 

25 

614 

619 

624 

628 

633 

638 

642 

647 

652 

656 

26 

661 

666 

670 

^75 

680 

685 

689 

694 

699 

7°3 

27 

708 

713 

717 

722 

727 

73i 

736 

741 

745 

750 

28 

755 

759 

764 

769 

774 

778 

783 

788 

792 

797 

29 

802 

806 

811 

816 

820 

825 

830 

834 

839 

844 

930 

848 

853 

858 

862 

867 

872 

876 

881 

886 

890 

31 

895 

900 

904 

909 

914 

918 

923 

928 

932 

937 

32 

942 

946 

95  * 

956 

960 

965 

970 

974 

979 

984 

4 

33 

988 

993 

997 

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0.4 
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34 

97035 

039 

044 

049 

053 

058 

063 

067 

072 

077 

3 

LJ.O 
1.2 

35 

081 

086 

090 

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104 

109 

114 

118 

123 

4 

1.6 

36 

128 

132 

*37 

142 

146 

!5* 

J55 

1  60 

165 

169 

5 

2.0 

37 

174 

179 

183 

1  88 

192 

197 

202 

206 

211 

216 

6 

2.4 

38 

220 

225 

230 

234 

239 

243 

248 

253 

257 

262 

7 

2.8 

39 

267 

271 

276 

280 

285 

290 

294 

299 

3°4 

308 

8 

3-2 

940 

3i3 

317 

322 

327 

331 

336 

340 

345 

350 

354 

9 

3-6 

41 

359 

364 

368 

373 

377 

382 

387 

39i 

396 

400 

42 

405 

410 

414 

419 

424 

428 

433 

437 

442 

447 

43 

45i 

456 

460 

465 

470 

474 

479 

483 

488 

493 

44 

497 

502 

506 

511 

516 

520 

525 

529 

534 

539 

45 

543 

548 

552 

557 

562 

566 

57i 

575 

580 

585 

46 

589 

594 

598 

603 

607 

612 

617 

621 

626 

630 

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635 

640 

644 

649 

653 

658 

663 

667 

672 

676 

48 

681 

685 

690 

695 

699 

704 

708 

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717 

722 

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727 

73i 

736 

740 

745 

749 

754 

759 

763 

768 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Pp.  Pts. 

XV111 


950-999 


N. 

0 

1 

2 

3 

4 

5 

6 

rr 

8 

9 

Pp.  Pts. 

950 

97  772 

777 

782 

786 

791 

795 

800 

804 

809 

813 

51 

818 

823 

827 

832 

836 

841 

845 

850 

855 

859 

52 

864 

868 

873 

877 

882 

886 

891 

896 

900 

905 

53 

909 

914 

918 

923 

928 

932 

937 

941 

946 

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54 

955 

959 

964 

968 

973 

978 

982 

987 

991 

996 

55 

98  ooo 

005 

009 

014 

019 

023 

028 

032 

037 

041 

56 

046 

050 

055 

059 

064 

068 

073 

078 

082 

087 

57 

091 

096 

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105 

109 

114 

118 

123 

127 

132 

58 

137 

141 

146 

150 

155 

J59 

164 

1  68 

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177 

59 

182 

1  86 

191 

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200 

204 

209 

214 

218 

223 

960 

227 

232 

236 

241 

245 

250 

254 

259 

263 

268 

61 

272 

277 

281 

286 

290 

295 

299 

3°4 

308 

313 

62 

318 

322 

327 

33i 

336 

340 

345 

349 

354 

358 

63 

363 

367 

372 

376 

381 

385 

390 

394 

399 

403 

5 

f\  f 

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408 

412 

417 

421 

426 

430 

435 

439 

444 

448 

2 

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65 

453 

457 

462 

466 

471 

475 

480 

484 

489 

493 

3 

1.5 

66 

498 

502 

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520 

525 

529 

534 

538 

4 

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67 
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543 

588 

547 

592 

552 
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610 

570 
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574 
619 

579 
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583 
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69 

632 

637 

641 

646 

650 

655 

659 

664 

66.8 

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677 

682 

686 

691 

695 

700 

704 

709 

713 

717 

8 

4.0 

71 

722 

726 

73i 

735 

740 

744 

749 

753 

758 

762 

9 

4-5 

72 

767 

771 

776 

780 

784 

789 

793 

798 

802 

807 

73 

811 

816 

820 

825 

829 

834 

838 

843 

847 

851 

74 

856 

860 

865 

869 

874 

878 

883 

887 

892 

896 

75 

900 

905 

909 

914 

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932 

936 

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76 

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963 

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972 

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77 

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994 

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069 

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79 

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083 

087 

092 

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114 

118 

980 

123 

127 

131 

136 

140 

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149 

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158 

162 

81 

167 

171 

176 

1  80 

185 

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193 

198 

202 

207 

4 

82 

211 

216 

220 

224 

229 

233 

238 

242 

247 

25  * 

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0.4 

83 

255 

260 

264 

269 

273 

277 

282 

286 

291 

295 

2 

0.8 

84 

300 

3°4 

308 

3*3 

317 

322 

326 

330 

335 

339 

3 

1.2 

85 

344 

348 

352 

357 

36i 

366 

370 

374 

379 

383 

4 

5 

1.6 

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86 

388 

392 

396 

401 

405 

410 

414 

419 

423 

427 

6 

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87 

432 

436 

441 

445 

449 

454 

458 

463 

467 

471 

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476 

480 

484 

489 

493 

498 

502 

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520 

524 

528 

533 

537 

542 

546 

550 

555 

559 

9 

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564 

568 

572 

577 

58i 

585 

590 

594 

599 

603 

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607 

612 

616 

621 

625 

629 

634 

638 

642 

647 

92 

651 

656 

660 

664 

669 

673 

677 

682 

686 

691 

93 

695 

699 

704 

708 

712 

717 

721 

726 

730 

734 

94 

739 

743 

747 

752 

756 

760 

765 

769 

774 

778 

95 

782 

787 

791 

795 

800 

804 

808 

813 

817 

822 

96 

826 

830 

835 

839 

843 

848 

852 

856 

86  1 

865 

97 

870 

874 

878 

883 

887 

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896 

900 

904 

909 

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9i3 

917 

922 

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935 

939 

944 

948 

952 

99 

957 

961 

965 

970 

974 

978 

983 

987 

991 

996 

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1 

2 

3   4 

5 

6 

7 

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9 

Pp.  Pts. 

THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 

AN  INITIAL  FINE  OF  25  CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY  AND  TO  $1.OO  ON  THE  SEVENTH  DAY 
OVERDUE. 


6 


1*48 


,**;  ., 


LD  21-100m-12,  '43  (8796s) 


VB   17267 


M306CH5 

QA 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


